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**Entry Task: Feb 12th Tuesday**

Define Buffer capacity You have 5 minutes

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**Agenda Discuss Buffer ws 1**

In-class notes little more buffer info and practice on Titrations HW: Buffers ws #2

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Explain why a mixture of HCl and KCl does not function as a buffer, whereas a mixture of HC2H3O2 and NaC2H3O2 does? HCl and KCl are conjugate-pairs, problem is that potassium is an alkali metal and will stay dissociated and add more + to the system making it more acidic.

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**2. What factors determine a) the pH, and b) the buffer capacity of a buffer solution?**

The pH of a buffer is determined by Ka for the conjugate acid present and the ratio of the conjugate base concentration to the conjugate acid concentration. The buffering capacity of buffer is determined by the concentrations of the conjugate acid and conjugate base present. Higher the concentration the higher the capacity.

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3. In a solution, when the concentrations of a weak acid and its conjugate base are equal, A) the system is not at equilibrium. B) the buffering capacity is significantly decreased. C) the -log of the [H+] and the -log of the Ka are equal. D) all of the above are true.

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4. Of the following solutions, which has the greatest buffering capacity? A) M HF and M NaF B) M HF and M NaF C) M HF and M NaF D) M HF and M NaF E) They are all buffer solutions and would all have the same capacity.

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5. The addition of hydrofluoric acid and __________ to water produces a buffer solution. A) HCl B) NaNO3 C) NaCl D) NaOH E) NaBr

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6. Which of the following could be added to a solution of sodium acetate to produce a buffer? A) acetic acid only B) acetic acid or hydrochloric acid C) hydrochloric acid only D) potassium acetate only E) sodium chloride or potassium acetate

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**7. A solution is prepared by dissolving 0**

7. A solution is prepared by dissolving 0.23 mol of hydrazoic acid and 0.27 mol of sodium azide in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the __________ present in the buffer solution. The Ka of hydrazoic acid is 1.9 × A) H2O B) H3O+ C) azide D) hydrazoic acid E) This is a buffer solution: the pH does not change upon addition of acid or base.

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**8. What is the pH of a buffer solution that is 0**

8. What is the pH of a buffer solution that is M in lactic acid and M in sodium lactate? The Ka of lactic acid is 1.4 × 10-4. Ka = [x][0.111] [0.211] Rearrange to get X by itself 1.4 x 10-4 = [x][0.111] [0.211] x= (1.4 x 10-4)(0.211) 0.111 3.57 x = [H+]= 2.66 x 10-4 pH = -log(2.66 x 10-4) = 3.57

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**Things we have not considered. volume changes. grams mole Molarity**

Things we have not considered *volume changes *grams mole Molarity *effect on pH *addition of acid/base and change in equilibrium

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**Buffer Problems Hurdles-**

1st -Which species in problem is “acid” or “base”? 2nd Kb value for OH- or Ka for H+ 3rd Setting up Ka or Kb expression correctly (1st) 4th Is there any changes in concentrations from given: *Grams Moles Molarity *Mixing two different volumes (M1V1 = M2V2) *Addition of a new species (ICE table)

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**Different Volumes Treat as dilution problems**

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**1a. Calculate the pH of a buffer that is 0. 100 M in NaHCO3 and 0**

1a. Calculate the pH of a buffer that is M in NaHCO3 and 0.125M in Na2CO3. b. Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3. Who is the conjugate in this reaction? HCO3- so we use H2CO3 Ka value= 5.6x10-11 Provide the Ka expression Provide the Ka express with # Ka = [H+][CO3-2] [HCO3-] 5.6 x = [x][0.125] [0.100] Rearrange to get X by itself x= (5.6 x 10-11)(0.100) 0.125 x = [H+]= 4.48 x 10-11 pH = -log(4.48 x 10-11) = 10.35

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**OR use the H-H equation [base] pH = pKa + log [acid] pH = 10.25 +**

1a. Calculate the pH of a buffer that is M in NaHCO3 and 0.125M in Na2CO3. b. Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3. Ka = [H+][CO3-2] [HCO3-] 5.6 x = [x][0.125] [0.100] OR use the H-H equation pH = pKa + log [base] [acid] pH = log [0.125] [0.100] pH = 0.0969 = 0.0969

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**1a. Calculate the pH of a buffer that is 0. 100 M in NaHCO3 and 0**

1a. Calculate the pH of a buffer that is M in NaHCO3 and 0.125M in Na2CO3. b. Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3. NOTICE we have different volumes!! What will be the TOTAL VOLUME? 120 mls We have “diluted” our mixture so we set up a dilution problem- MAKE SURE volume units are the same!! M1V1 = M2V2 (0.20M)(0.055L) = (x) (0.120L)= 9.17x10-2 M of NaHCO3 (new M) (0.15M)(0.065L) = (x)(0.120L) = 8.125x10-2 Mof Na2CO3 (new M) x= (5.6 x 10-11)(9.17x10-2) 8.125x10-2 x = [H+]= 6.32 x 10-11 pH = -log(6.32 x 10-11) = 10.20

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**1a. Calculate the pH of a buffer that is 0. 100 M in NaHCO3 and 0**

1a. Calculate the pH of a buffer that is M in NaHCO3 and 0.125M in Na2CO3. b. Calculate the pH of a solution formed by mixing 55mls of 0.20M NaHCO3 with 65mls of 0.15 M Na2CO3. NOTICE we have different volumes!! What will be the TOTAL VOLUME? 120 mls H-H equation 5.6 x = [x][8.125x10-2] [9.17x10-2] pH = log [8.125 x10-2] [9.17x10-2 10.20 =

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**[base] pH = pKa + log [acid] 3.8 = 3.85 + -0.0378**

2a. Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. b. Calculate the pH of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. Ka= 1.4 x 10-4 = [x][0.11] [0.12] pH = pKa + log [base] [acid] pH = log [0.11] [0.12] 3.8 =

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**2a Calculate the pH of a buffer that is 0. 12 M in lactic acid and 0**

2a Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11M in sodium lactate. B. Calculate the pH of a solution formed by mixing 85mls of 0.13M lactic acid with 95mls of 0.15M sodium lactate. What will be the TOTAL VOLUME? 180 mls We have “diluted” our mixture so we set up a dilution problem- MAKE SURE volume units are the same!! (0.13M)(0.085L) = (x)(0.180L) = 6.14x10-2M of lactic acid (new M) (0.15M)(0.095L) = (x)(0.180L)= 7.92x10-2 M of sodium lactate (new M) pH = log [7.92 x10-2] [6.14x10-2] 3.96 = 0.011

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**Addition of strong acid-base**

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**Addition of strong acid**

Using the same information given in 1a calculate the pH after the addition of 0.010M of HCl. NaHCO3 + HCl H CO3-2 NaHCO3 H+ CO3-2 Before 0.100 mol 0.010 mol 0.125 mol Change mol -0.010 mol After Reaction 0.110 mol 0.000 mol 0.115 mol Added neutralizes any CO3-2 ions

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**Calculating pH Changes in Buffers**

Ka expression with NEW molarities to calculate new pH Ka = [H+][CO3-2] [HCO3-] 5.6 x = [x][0.115] [0.110] Rearrange to get X by itself x= (5.6 x 10-11)(0.110) 0.115 x = [H+]= 5.83 x 10-11 pH = -log(3.97 x 10-11) = 10.3

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**Calculating pH Changes in Buffers**

Using the same information given in 1a calculate the pH after the addition of 0.050M of NaOH NaHCO3 OH− CO2−2 Before 0.100 mol 0.050 mol 0.125 mol Change mol -0.050 mol After Reaction 0.000 mol 0.175 mol Added neutralizes any H+ ions

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**Calculating pH Changes in Buffers**

Ka expression with NEW molarities to calculate new pH Ka = [H+][CO3-2] [HCO3-] 5.6 x = [x][0.175] [0.050] Rearrange to get X by itself x= (5.6 x 10-11)(0.050) 0.175 x = [H+]= 1.6 x 10-11 pH = -log(1.6 x 10-11) = 10.8

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Titration: A laboratory method for determining the concentration of an unknown acid or base using a neutralization reaction. A standard solution,(a solution of known concentration-titrant), is used.

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**(M1)(V1). The unknown has a known volume(V2) so we can calculate (M2).**

Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base). After titration- we have a known volume and concentration from titrant (M1)(V1). The unknown has a known volume(V2) so we can calculate (M2). (M1)(V1)= (M2)(V2) Titrant-known of concentration Unknown concentration with a known volume

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Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

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pH Titration Curve pH titration curve, a graph of pH as a function of volume of the added titrant. The pH curve can: *help determine equivalence point *determine the pH indicators needed for Ka or Kb determination.

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**pH Titration Curve Can you identify the titrant?**

Is it an acid or base. Base *pH is increasing *it levels off with lots of base.

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**Strong Acid- Strong Base Titration 4 regions of a titration curve**

1st- Initial pH- its really low- probably a strong acid. 2nd- Between initial and equivalence pt. rises slowly then rapidly around the ~SAME~ volume as the unknown.

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**Strong Acid- Strong Base Titration 4 regions of a titration curve**

3rd Equivalence pt [H+] = [OH-] = pH 7 4th After equivalence pt. Has plateaued with excess base

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**Titration of a Strong Acid with a Strong Base**

From the start of the titration to near the equivalence point, the pH goes up slowly.

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**Titration of a Strong Acid with a Strong Base**

Just before and after the equivalence point, the pH increases rapidly.

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**Titration of a Strong Acid with a Strong Base**

At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

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**Titration of a Strong Acid with a Strong Base**

As more base is added, the increase in pH again levels off.

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**We will learn how to CALCULATE the equivalence point later.**

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**Titration of a Weak Acid with a Strong Base**

Which way will the equilibrium shift in the case of weak acid and strong base?

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**Titration of a Weak Acid with a Strong Base**

1st- Initial pH- ~3 or 4 is a “stronger” weak acid 2nd- Between initial and equivalence pt. 2 things to consider 1. neutralization of weak acid by strong base 2. Strong base acts as a buffer so it resists the titration pH of your weak acid is ½ the amount for neutralization of base- ½ of 50 mls (25 mls) and pH of ~4.8

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**Titration of a Weak Acid with a Strong Base**

3rd At equivalence pt ~The pH here is above 7 which is what we expected 4th After equivalence pt The curve looks very similar to a strong acid/strong base curve.

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**Titration of a Weak Base with a Strong Acid**

Which way will the equilibrium shift in the case of weak base and strong acid?

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**Titration of a Weak base with a Strong acid**

Strong base 1st- Initial pH- its really high probably a strong base. 2nd- Between initial and equivalence pt. 2 things to consider 1. neutralization of weak base by strong acid 2. Strong acid acts as a buffer so it resists the titration

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**Titration of a Weak base with a Strong acid**

Strong base 3rd At equivalence pt ~The pH here below 7 which is what we expected 4th After equivalence pt The curve looks very similar to a strong acid/strong base curve.

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**Titration of a Weak Acid with a Strong Base**

With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

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**Titrations of Polyprotic Acids**

In these cases there is an equivalence point for each dissociation.

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