Buffers Made Easy.

Buffers Made Easy

What is a Buffer? HA ⇋ H+ + A− X-NH2 + H2O  X-NH3+ + OH−
An aqueous solution that has a highly stable pH even upon addition of small amounts of H+ or OH−. The solution is able to absorb and neutralize incoming acid or incoming base. How?? Because of the presence of significant quantities of both a conjugate weak acid/base pair. HA ⇋ H+ + A− X-NH2 + H2O  X-NH OH−

There are two methods to construct a buffer
Dissolve a conjugate WA-salt (or WB-salt) into a solution of WA or WB. Titrate a WA (or WB) partway along. HA ⇋ H+ + A− NH3 + H2O  NH OH−

The pH would stay the same
The pH of an acetic acid solution is ~ 4.7 If you added some sodium acetate to acetic acid, The pH would go up The pH would go down The pH would stay the same

HA ⇋ H+ + A− HA ⇋ H+ + A− The pH of an acetic acid solution is ~ 4.7
If you added some sodium acetate, A− to acetic acid, The pH would go up HA ⇋ H+ + A− Adding A− will shift the equilibrium to the left, reducing the amount of H+. This is the common ion effect The pH would go down The pH would stay the same

The pH would stay the same
The pH of an acetic acid solution is ~ 4.7 If you added some sodium hydroxide to acetic acid, The pH would go up The pH would go down The pH would stay the same

The pH of an acetic acid solution is ~ 4.7
HA ⇋ H+ + A− If you added some sodium hydroxide, OH− to acetic acid, The pH would go up HA ⇋ H+ + A− + H2O LeChatelier’s Principle, adding OH− will combine with H+ forming water, and shift the equilibrium to the right. As the becomes larger, due to the “shift right,” the the [H+] to be smaller The pH would go down The pH would stay the same

What is the pH of a solution made with equal moles of HC2H3O2 and C2H3O2−? Ka = 1.8 × 10-5

When WA = WB in a buffer pH = pKa pH = 1.80
What is the pH of a solution made with equal moles of HC2H3O2 and C2H3O2−? Ka = 1.8 × 10-5 pH = 9.26 pH = 7.00 pH = 4.74 When WA = WB in a buffer pH = pKa pH = 1.80

There are no spectator ions. Cl– Both NH4+ and Cl– NH4+
A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which of the ion(s) is(are) a spectator ion(s) and not involved in the equilbrium? There are no spectator ions. Cl– Both NH4+ and Cl– NH4+

There are no spectator ions. Cl– Both NH4+ and Cl– NH4+
A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which of the ions is a spectator ion? There are no spectator ions. Cl– Both NH4+ and Cl– NH4+

NH3(aq) + H2O(L)  NH4+(aq) + OH–(aq)
A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which equilibrium equation(s) accurately describes the reaction to allow for pH calculation? NH3(aq) + H2O(L)  NH4+(aq) + OH–(aq) NH4+(aq) + OH–(aq)  NH3(aq) + H2O(L) H3O+(aq) + OH–(aq)  2H2O(l) NH4+(aq) + H2O(L)  H3O+(aq) + NH3(aq) H3O+(aq) + NH3(aq)  NH4+(aq) + H2O(L)

A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Which equilibrium equation(s) accurately describes the reaction to allow for pH calculation? NH3(aq) + H2O(L)  NH4+(aq) + OH–(aq) NH4+(aq) + H2O(L)  H3O+(aq) + NH3(aq)

A mixture of 0.10 mol of NH4Cl and 0.12 mole of NH3 is added to enough water to make 1.0 L of solution. Now, add 250 ml of 0.20 M HCl and calculate the pH. NH3(aq) + H2O(L)  NH4+(aq) + OH–(aq) NH4+(aq) + H2O(L)  H3O+(aq) + NH3(aq)

Which of the following combinations will result in a buffer?
Will the pH of the solution be 7, above 7, or below 7? 20 ml of 1 M KNO3 and 20 ml of 1 M HNO3 A solution of calcium acetate and acetic acid 100 ml of 0.5 M CH3NH2 with 5 g of dissolved CH3NH3NO3 (MM = 94) 20 ml of 1 M NaOH and 20 ml of 1 M NaCl

Will the pH of the solution be 7, above 7, or below 7? 20 ml of 1 M KNO3 and 20 ml of HNO3 SA (HNO3) and it’s pathetic (NO3−) does NOT make a buffer A solution of calcium acetate and acetic acid? WA (acetic acid)and it’s conj WB (acetate ion) = buffer, pH < 7 100 ml of 0.5 M CH3NH2 with 5 g of dissolved CH3NH3NO3 WB (CH3NH2) and it’s conj WA (CH3NH3+) = buffer, pH > 7 20 ml of 1 M NaOH and 20 ml of NaCl SB (NaOH) and it’s pathetic (NO3−) does NOT make a buffer

Will the pH of the solution be 7, above 7, or below 7? 50 ml of 0.5 M NaOH with 25 ml of 0.5 M HF 20 ml of 1 M HNO3 and 10 ml of 1 M NaOH 100 ml of 0.5 M CH3NH2 with 100 ml of 0.5 M HNO3 100 ml of 0.5 M CH3NH2 with 60 ml of 0.5 M HNO3

Will the pH of the solution be 7, above 7, or below 7? 50 ml of 0.5 M NaOH with 25 ml of 0.5 M HF WA (HF) titrated beyond equiv point = not buffer, pH > 7 20 ml of 1 M HNO3 and 10 ml of 1 M NaOH SA (HNO3) and SB (NaOH) in any quantities does NOT make a buffer 100 ml of 0.5 M CH3NH2 with 100 ml of 0.5 M HNO3 WB (CH3NH2) titrated to equiv point does not make a buffer 100 ml of 0.5 M CH3NH2 with 60 ml of 0.5 M HNO3 WB (CH3NH2) titrated partway = buffer, likely pH > 7

adding HCl adding KOH adding H2O adding NaNO2
Consider the weak acid ionization reaction represented below. HNO2 ⇋ H+ + NO2− Which of the actions below would shift the equilibrium position toward reactants? adding HCl adding KOH adding H2O adding NaNO2

common ion H+ will shift equilibrium to the left adding KOH adding H2O
Consider the weak acid ionization reaction represented below. HNO2 ⇋ H+ + NO2− Which of the actions below would shift the equilibrium position toward reactants? adding HCl common ion H+ will shift equilibrium to the left adding KOH adding H2O adding NaNO2

Calculate the pH of a 0.025 M HNO2 solution. Ka = 4.5 x 10−4

Calculate the pH of a 0.025 M HNO2 solution. Ka = 4.5 x 10−4
For the reaction: HA ⇋ H+ + A− this is an x2 problem

Calculate the pH of a solution made by adding 2
Calculate the pH of a solution made by adding 2.1 g of sodium nitrite to 2 L of M HNO2 solution. Ka = 4.5 x 10−4 (Assume no volume change when the salt is added.) MM = 69 g/mol

This is a buffer solution
Calculate the pH of a solution made by combining 0.20 mole of sodium nitrite with 0.20 mole of HNO2 in a 2.0 L solution. Ka = 4.5 x 10−4 This is a buffer solution

HCl / NaCl HC2H3O2 / NH3 H3PO4 / NaH2PO4 HNO3 / Ca(OH)2 HCN / NaOH
Which of the following combinations can produce a buffer solution? Select all that apply. HCl / NaCl HC2H3O2 / NH3 H3PO4 / NaH2PO4 HNO3 / Ca(OH)2 HCN / NaOH NH4NO3 / NH3

Which of the following will produce a buffer solution?
HCl / NaCl HC2H3O2 / NH3 This solution can buffer, with a WA and WB, but since they are not conjugates...you will never see them together in a question. H3PO4 / NaH2PO4 H3PO4− ⇋ H+ + H2PO42− HNO3 / Ca(OH)2 HCN / NaOH Acid with ~half base NH4NO3 / NH3 Two ways to make a buffer put in a conjugate WA /WB pair put a WA with ~half of SB or WB with ~half of SA

KNO3 / NaOH HCN / NaOH NH3 / HCl HCl / KOH NH3NO3 / NH3
Which of the following combinations can produce a buffer solution? Select all that apply. KNO3 / NaOH HCN / NaOH NH3 / HCl HCl / KOH NH3NO3 / NH3

Which of the following will produce a buffer solution?
KNO3 / NaOH HCN / NaOH NH3 / HCl HCl / KOH NH4NO3 / NH3 Two ways to make a buffer put in a conjugate WA /WB pair put a WA with ~half of SB or WB with ~half of SA

Cannot be determined without knowing volumes of each, or total volume
What is [H+] of a M HF solution dissolved with enough solid NaF to produce a solution 0.20 M NaF Ka (HF) = 6.8 x 10−4 1.7 x 10-4 M 3.4 x 10-4 M 6.8 x 10-4 M 1.4 x 10-3 M 2.7 x 10-3 M Cannot be determined without knowing volumes of each, or total volume

Cannot be determined without knowing volumes of each, or total volume
What is [H+] of a M HF solution dissolved with enough solid NaF to produce a solution 0.20 M NaF Ka (HF) = 6.8 x 10−4 1.7 × 10-4 M 3.4 × 10-4 M 6.8 × 10-4 M 1.4 × 10-3 M 2.7 × 10-3 M Cannot be determined without knowing volumes of each, or total volume

Calculate the pH of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in 0.50 M acetic acid (HC2H3O2 or HAc) pKa (HAc) = 4.74 3.74 4.24 4.74 5.24 5.74

For any buffer, when [WA] = [WB], the pH = pKa
Calculate the pH of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in 0.50 M acetic acid (HC2H3O2 or HAc). pKa (HAc) = 4.74 3.74 4.24 4.74 5.24 5.74 For any buffer, when [WA] = [WB], the pH = pKa

Using equal volumes and concentrations of which of the following combinations would be best to make a buffer with a pH above 7 ? NaCl and HCl NH3 and NH4Cl NaOH and HCl HNO2 and NaNO2 NH3 and C5H5NHCl acid pKa HNO2 3.35 C5H5NH2+ 5.23 NH4+ 9.25

Using equal volumes and concentrations of which of the following combinations would make a buffer with a pH above 7 ? NaCl and HCl NH3 and NH4Cl NaOH and HCl HNO2 and NaNO2 NH3 and C5H5NHCl When the conjugate pair [base] = [acid] in the above expression Ka = [H+], so pKa = pH acid pKa HNO2 3.35 C5H5NH2+ 5.23 NH4+1 9.25

Calculate the pH of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in M acetic acid (HC2H3O2 or HAc) pKa (HAc) = 4.74 3.74 4.24 4.74 5.24 5.74

Note that the pH was diffeent from pKa by one unit
Calculate the pH of a 0.50 M solution of sodium acetate (NaC2H3O2 or NaAc) in M acetic acid (HC2H3O2 or HAc) pKa (HAc) = 4.74 3.74 4.24 4.74 5.24 5.74 Note that the pH was diffeent from pKa by one unit

No Changes because OH− is not part of the equilibrium reaction.
In an NH4+/NH3 buffer, what concentration changes will occur if a small amount of OH− is added? [NH4+] [NH3] pH 1 Increase 2 Decrease 3 4 5 6 No Changes because OH− is not part of the equilibrium reaction.

No Changes because OH-1 is not part of the equilibrium reaction.
In an NH4+/NH3 buffer, what concentration changes will occur if a small amount of OH- is added? [NH4+] [NH3] pH 1 Increase 2 Decrease 3 4 5 6 No Changes because OH-1 is not part of the equilibrium reaction. A shift will always occur, however it may not always make any appreciable effect on the pH value. To think about this write the equilibrium reaction (from either perspective) NH3 + H2O → OH− + NH4+ adding OH− shifts left OR NH4+ → NH3 + H+ adding OH− shifts right

The pKa of phenolphthalein is likely to be HIn ⇋ H+ + In−
below 8 ~8 ~9 ~10 above 10 there is not enough information to know

Calculate the pH of a buffer solution containing 0
Calculate the pH of a buffer solution containing 0.20 moles sodium acetate and 0.30 moles acetic acid (HAc) pKa (HAc) = 4.74. 2.63 4.57 4.74 4.92 5.27 not enough information, volume information must be known to calculate a pH. Yes, you may use a calculator

Calculate the pH of a buffer solution containing 0
Calculate the pH of a buffer solution containing 0.20 moles sodium acetate and 0.30 moles acetic acid (HAc) pKa (HAc) = 4.74. 2.63 4.57 4.74 4.92 5.27 not enough information, volume information must be known to calculate a pH.

HCHO2/CHO2− HCO3−/CO3-2 HNO3/NO3−
Which of the following conjugate acid-base pairs will not function as a buffer? HCHO2/CHO2− HCO3−/CO3-2 HNO3/NO3− All acid-base pairs will function as buffers. HCHO2 and CHO2– will not work as a buffer because CHO2– is a spectator ion. HCO3– and CO32– will not work as a buffer because HCO3– is a spectator ion. HNO3 and NO3– will not work as a buffer because NO3– is a spectator ion.

HCHO2/CHO2-1 HCO3-1/CO3-2 HNO3/NO3-1
Which of the following conjugate acid-base pairs will not function as a buffer? HCHO2/CHO2-1 HCO3-1/CO3-2 HNO3/NO3-1 All acid-base pairs will function as buffers. HCHO2 and CHO2– will not work as a buffer because CHO2– is a spectator ion. HCO3– and CO32– will not work as a buffer because HCO3– is a spectator ion. HNO3 and NO3– will not work as a buffer because NO3– is a “pathetic” spectator ion.

not enough information, total volume must be known.
Calculate the pH of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HAc) to which 0.20 moles HCl are added pKa (HAc) = 4.74. 3.74 4.24 4.74 5.24 5.74 not enough information, total volume must be known.

Calculate the pH of 2000 ml of a buffer solution containing 0
Calculate the pH of 2000 ml of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HAc) to which 0.35 moles HCl are added pKa (HAc) = 4.74 submit your answer.

This is a buffer which has been pushed beyond its buffer capacity.
Calculate the pH of 2000 ml of a buffer solution containing 0.25 moles sodium acetate and 0.30 moles acetic acid (HAc) to which 0.35 moles HCl are added pKa (HAc) = 4.74 This is a buffer which has been pushed beyond its buffer capacity. The 0.25 moles of acetate ion would be able to neutralize 0.25 moles of incoming HCl, however that would leave 0.10 moles of excess HCl. This would be more important than any H+ contributed by the weak acid.

Predicting Strength of Acids
Using the Chemical Formula to Predict Acid Strength

Three Groups of Acids In which we will look for trends
Oxyacids HNO3, HNO2 H2SO4, H2SO3 HClO HClO4 Binary acids, H-X HF, HCl, HBr, HI H2O, H2S Carboxylic Acid (organic acid, containing double bonds near the ionizable H+) CxHyOz −COOH

3 Factors causing stronger acids
Polarizability of the −H bond Weakness of the X−H bond Stability of the conjugate base (the resulting anion).

Which acid is the weakest?
HBrO HBrO2 HBrO3 HBrO4

HBrO HBrO2 HBrO3 HBrO4 Oxy acids with varying # of O atoms more oxygens = stronger acid Less oxygens have less electron drawing capability, thus the -H bond is less polarized, thus a weaker acid. Also, the additional oxygens help to stabilize the conjugate base by providing more room to “spread out” its negative charge.

Oxyacids: HXOn (really XOnH)
As additional oxygens attached to X, increases acidity. HClO < HClO2 < HClO3 < HClO4 The additional oxygens “draw” the electrons in the H−O bond causing the bond to become even more polarized. Also, the additional oxygens help to stabilize the conjugate base by providing more room to “spread out” its negative charge.

HClO2 HBrO2 HIO2 H−O−Cl−O H−O−Br−O H−O−I−O

H−O−Cl−O HClO2 HBrO2 HIO2 Oxy acids different central atom but same # oxygens I is less electronegative than Cl thus the −H bond is less polarized, thus the H is more likely to “hang on” and a therefore a weaker acid. The resulting conjugate base ion will be less stable since the presence of the less electronegative I atom is not able to distribute and stablilize the negative charge as well as a more electronegative atom. H−O−Br−O H−O−I−O

Oxyacids: HOX As the electronegativity of X increases, the acidity increases HOI < HOBr < HOCl Because the H−O bond becomes more polarized due to greater “electron-drawing” of the nearby electronegative atom. The resulting conjugate base ion will be more stable since the presence of the more electronegative Cl atom can help distribute and stablilize the negative charge.

Which molecule is the weakest acid?
H3P H2S HCl

H3P H2S HCl Binary acids across a row P is less electronegative than Cl thus the −H bond is less polarized, thus the H is more likely to “hang on” and a weaker acid results.

H2S H2Se H2Te

H2S H2Se H2Te Binary acids in a column S is smaller in size and thus the bond is stronger resulting in a weaker acid. (If you use the electronegativity argument, you will get to the wrong answer.)

Making Sense of Binary Acid Strength
What is the explanation.....?

Binary Acids: HX Down a column, bond strength is the issue (or weakness) is more important: H2O < H2S < H2Se As X’s atomic size gets larger, bond length is longer, making the bond not as tight, causing acid strength to increase (stronger acid). size changes dramatically down the column the H−X bond strength decreases, because the X atoms electrons are more diffuse (spread over a larger electron cloud) and attracts the H+ ion less effectively.

Binary Acids: HX Across a row, bond polarity is more important: NH3 < H2O < HF As the electronegativity of X increases, size does change acrosse the row, but the changes are far less dramatic across the row and thus have less of an effect of acidity. the polarity of bond increases, causing the acid strength to increase.

Carboxylic Acids: stuff−COOH
Stonger acid caused by more stable conjugate After the H+ falls off, the C=O and C−O bonds can exhibit resonance Resonance is the relocation electrons in an available nearby location resulting in an “averaging” of the structures. Resonance allows the electrons to “spread out,” distributing the charge and stabilizing the conjugate base. H

Carboxylic Acids: Stonger acid caused by
more polar −H bond and more stable conjugate Acidity increases as more electronegative atoms are attached to the root of the acid because of increased electron-drawing capability and greater polarization of the O−H bond. because the more electronegative F atoms the presence of the more electronegative F atom can help distribute and stablilize the negative charge. < F CH3COOH < CF3COOH

What Causes H+’s to Fall Off?
Acid is only pH changing when H+s fall off Not every H falls off Not every H attached to an O falls off When the H falls off, the conjugate base is quite stable. What makes this ion stable? the resonance that can occur with the nearby C=O bond. Spreads out (smears) the negative charge over more atoms

Notable NON-acids Why is stuff−COOH an acid,
yet stuff−COH and metal−OH is NOT? stuff−COH is an alcohol without any nearby electron drawing C=O or C=C groups, the O−H bond is not polarized enough. metal−OH is a hydroxide with an ionic bond that is “super” polarized = ionic, thus metal−OH is a base because of the hydroxide pushed into the water.

Acid Strength Summary When more oxygens attached to X
Oxygens draw electrons away from −O−H bond = higher acidity because −O−H bond becomes more polarized the resulting ion is larger and the − charge can spread out When there are same # oxygens more electronegative X = higher acidity the resulting ion has a more electronegative ion which is better at spreading out (and stabilizing) the negative charge Binary acids acids in a row because −X−H bond becomes more polarized, and H’s falls of easier Binary acids in a column it’s not about electronegativy Larger radius X, weaker bond = higher acidity

Which of the following binary acids is the weakest acid in water
H2S H2Se H2Te HBr HI

Which of the following binary acids is the weakest acid in water
H2S H2Se H2Te HBr HI We know the answer is not HBr and HI, which are both nearly completely ionized in water. Consider the 2− ions that can be formed from the other acids. S2−, Se2−, and Te2−. Based on higher density of negative charge, the sulfide ion is the strongest of these bases. Therefore its conjugate acid must be the weakest, leading to the conclusion that H2S is the weakest of these acids. For those who wish to memorize mindless rules, the strenghs of binary acids increases to the right and down on the periodic table, with HI being the strongest, and H2S the weakest.

Which weak acid is the weakest? Do not use your Ka chart.
hypochlorous acid hypobromous acid hypoiodous acid these weak acids are all equally weak there is not enough information to distinguish these are not weak acids, they are all strong acids

Which weak acid is the weakest? Do not use your Ka chart.
hypochlorous acid hypobromous acid hypoiodous acid Since these acids all have the same number of oxygens, the distinguishing feature is the electronegativity of the atom adjacent to the ionizing -OH bond. Since I is the least electronegative, it polarizes the -OH bond least making for the weakest acid.

Which of the following acids can be oxidized to form a stronger acid
Which of the following acids can be oxidized to form a stronger acid? (Select all that apply.) H2SO4 HNO2 HBrO2 HIO3 HClO4 H3PO4

Which of the following acids can be oxidized to form a stronger acid?
H2SO4 HNO2 HBrO2 HIO3 Any acid to which more oxygens can be added must attain a higher oxidation number to add those oxygens and thus would become oxidized HClO4 H3PO4

HClO4 HClO3 HBrO3 HBrO4 HIO3 HIO4

HClO4 HClO3 HBrO3 HBrO4 HIO3 HIO4 Oxy acids: varying oxygens and varying central atom looking for less oxygens and less electronegative central atom AP would not force you to decide which factor trumps the other: such as HBrO2 vs HIO3.

Which base is the strongest?
ClO4- ClO3- BrO3- BrO4- IO3- IO4-

Which base is the strongest?
ClO4- ClO3- BrO3- BrO4- IO3- IO4− The weaker the WA, then the stronger the cjWB

chlorous nitrous acetic hypobromous they are all equally weak there is not enough information to distinguish acid pKa chlorous 1.96 nitrous 3.35 acetic 4.74 hypobromous 8.6

chlorous nitrous acetic hypobromous they are all equally weak there is not enough information to distinguish The largest pKa is the weakest acid acid pKa chlorous 1.96 nitrous 3.35 acetic 4.74 hypobromous 8.6

What Causes H+ ’s to Fall Off?
Acid is only pH changing when H+ ’s fall off Not every H falls off Not every H attached to an O falls off When the H falls off, the conjugate base is quite stable. What makes this ion stable? the resonance that can occur with the nearby C=O bond. Spreads out (smears) the negative charge over more atoms

Polyprotic Acids H2A and H3A

Polyprotic Acids Acids with more than one ionizable proton
The second H+ ions never ionize as easily as the first. After the first ionization, the H+ ion does not leave the resulting negatively charged ion as easily. Successive Ka values become increasingly smaller. If the Ka values differ by a factor of 103 or more, the pH can be calculated by considering only the first ionization. Subsequent ionizations contribute a negligible amount of H+ ions.

Examples of Polyprotic Acids

Titration of Polyprotic Acids
Phosphoric Acid Titration

Brønsted & Lowry Definition as applied to a strong acid
HCl(g) H2O(L) → H3O Cl− HCl is acting as the proton donor. H2O is acting as the proton acceptor. But for a strong acid, this B & L definition is not necessary, it’s just as convenient and accurate to use the Arrhenius definition that defines an acid as a substance that dissolves to produce H+ HCl(aq) → H Cl−

Brønsted & Lowry Definition as applied to a strong base
NaOH(s) + H2O(L) → Na+ + OH− + H2O(L) This of course fits the Arrhenius definition by dissolving to produce OH− So what about the proton donor and acceptor? We could say that water donates a proton, and the OH− accepts a proton. OH− H2O(L) → H2O(L) OH− H2O(L) is acting as the proton donor. OH− is acting as the proton acceptor. But of course it seems silly to even write this out because the exchange of proton turns into exactly the same thing. So for a strong base, this B & L definition is also not necessary, its just as convenient and accurate to use the Arrhenius definition that defines a base as a substance that dissolves to produce OH−.

Let’s Define Acids & Bases
In 1880’s Arrhenius (same guy who studied reaction rates) defined: Acid: a substance that dissolves in water to form H+ HCl → H+ + Cl− Base: a substance that dissolves in water to form OH− NaOH → Na+ + OH− 1923 Brønsted & Lowry defined: Acids as a proton (H+) donor Bases as a proton (H+) acceptor Also in 1920’s Gilbert Lewis (same guy who invented {Electron Dot} Lewis Structures) expanded the definition: Acids as electron pair acceptors Bases as electron pair donors We have discussed this in reference to ions solvated in aqueous solution maybe more on this later...

Acids and Bases as gases?
Most reactions we will study will be in solution, but gases can also act as acids and bases. You may recall... We tested the speed of gases with the following reaction: HCl(g) + NH3(g) → NH4Cl(s) Why was the white ring NOT in the middle? Why was it closer to the HCl end? NH4Cl NH4Cl forms

Most Acids and Bases will be in an aqueous solution, however they can occur in the gas phase as well. What makes the reaction on the previous slide and acid base reaction? HCl(g) + NH3(g) → NH4Cl(s) Draw a Lewis Structure the following particles HCl NH3 NH4+ Cl-

Most Acids and Bases will be in an aqueous solution, however they can occur in the gas phase as well. What makes the reaction on the previous slide and acid base reaction? HCl(g) + NH3(g) → NH4Cl(s) Draw a Lewis Structure the following particles HCl NH3 NH4+ H+

Buffers Made Easy.

Similar presentations

Presentation on theme: "Buffers Made Easy."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Buffers Made Easy.

Similar presentations

Presentation on theme: "Buffers Made Easy."— Presentation transcript:

Similar presentations

About project

Feedback