Acids and Bases

Properties of Acids and Bases Pg 236

Why do Acids and bases change ACIDSBASES Sour taste (vinegar)Bitter taste (baking soda) React with some metal to form H 2 gas (Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) Feels slippery (soap) Turns blue litmus redTurns red litmus blue

Arrhenius Definition Acid: produces H + (or H 3 O + ) when dissolved in water HCl(aq)  H + (aq) + Cl - (aq) H + (aq) + H 2 O(l)  H 3 O + (aq) NOTE: H 3 O + = hydronium ion Base: produces OH - when dissolved in water. NaOH(s)  Na + (aq) + OH - (aq)

Bronsted-Lowry Definition Acids: proton (H + ) donors HF(aq)  H + (aq) + F - (aq) H + (aq) + H 2 O(l)  H 3 O + (aq) Bases: proton acceptors NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) H 2 O: acts as an acid and a base = AMPHOTERIC

Strength of Acids Strong Acids:  ionize (splits up into ions) almost 100% in water  mostly ions in solution  amount of HCl present is negligeable HCl(aq)  H + (aq) + Cl - (aq) Weak acids:  ionize poorly in water  not many of these ions present in solution  mostly acetic acid (HC 2 H 3 O 2 ) HC 2 H 3 O 2 (aq) C 2 H 3 O 2 - (aq) + H + (aq) NOTE: strong acids are strong electrolytes and will conduct electricity better than weak acids.

Strength of Bases Strong Bases: ionize almost 100% in water NaOH(s)  Na + (aq) + OH - (aq) Weak Bases: ionize poorly in water NH 3 (l) + H 2 O(l) NH 4 + (aq) + OH - (aq) NOTE: strong bases are strong electrolytes

Conjugate Acids and Conjugate Bases these differ by only one proton Examples HCl Cl - SO 4 2- HSO 4 - Lose a proton Gain a proton Acid Conjugate base of HCl BaseConjugate acid of SO 4 2-

Reactions with Water Conjugate acid-base pair: CH 3 CO 2 H/CH 3 CO 2 - Conjugate acid-base pair: H 2 O/H 3 O +

Monoprotic, Diprotic and Triprotic Monoprotic  donates one acidic proton  eg: HCl + H 2 O  H 3 O + + Cl -  only one H + to donate Diprotic  donates two acidic protons  eg: H 2 SO 4 + H 2 O  H 3 O + + HSO 4 -  HSO H 2 O  H 3 O + + SO 4 2- Triprotic  donates three acidic protons  eg: H 3 PO 4 + H 2 O  H 3 O + + H 2 PO 4 -  three H + to donate

Homework Pg 251 #1, 2 Pg 253 #4, 5, 6

pH < 7 acidic pH = 7 neutral pH > 7 basic

 a measure of acid strength  By definition all acids contain at least one acidic proton = H +  HA is a symbol used to represent any general acid HA H + (aq) + A - (aq) H + + H 2 O  H 3 O +   [H + ] = [H 3 O + ]  If a lot of H 3 O + is produced the solution is very acidic.  pH is directly related to [H 3 O + ]. pH

H 2 O(l) + H 2 O(l)  H 3 O + (aq) + OH - (aq) This reaction does not occur to any great extent. [H 3 O + ] = 1 x mol/L [OH - ] = 1 x mol/L Because both concentrations are equal water is said to be neutral. Therefore, if [H 3 O + ] = [OH - ] neutral [H 3 O + ] > [OH - ] acidic [H 3 O + ] < [OH - ] basic NOTE: [H 3 O + ][OH - ] = 1.0 x Self-Ionization of Water

Expressing hydronium concentrations in scientific notation isn’t very convenient. The pH scale was developed to make the expression of H 3 O + concentration more convenient. [H 3 O + ] is the concentration in mol/L pH = -log[H 3 O + ]

The concentration of H 3 O + is 1.0 x Calculate the pH. Example 1: pH of Water pH = -log[H 3 O + ] = -log(1.0 x ) = -(-7) = 7 The pH of water is 7. Therefore pH 7 is neutral.

Determine the pH of a 1M solution of HCl. Example 2 HCl (aq)  H + + Cl - 1M x Therefore [H 3 O + ] = 1 Therefore, pH = -log[H 3 O + ] = -log(1) = 0 Therefore a 1M solution of HCl has pH 0.

What is the pH of a 0.01M solution of HCl? Example 3 HCl (aq)  H + + Cl - [H 3 O + ] = 0.01 M Therefore, pH = -log[H 3 O + ] = -log(0.01) = 2

What is the pH of a 1M NaOH solution? Example 4 pOH = -log[OH - ] = -log(1) = 0 pH + pOH = 14 pH = 14 – pOH pH = 14 Therefore a 1M solution of NaOH has pH 14. The pH of a very basic solution.

Determine the pH of a 0.01M NaOH solution. Example 5 pOH = -log[OH - ] = -log(0.01) = 2 pH + pOH = 14 pH = 14 – pOH pH = 12 Therefore a 1M solution of NaOH has pH 12. The pH of a basic solution.

The pH reading of a solution is What is its hydrogen ion concentration? Example pH = [H + ] = [H + ] 4.7  mol/L = [H + ] Base ten logarithm represents an exponent log 10 (100) =2 10 2

Calculate the pH of a M H 2 SO 4 solution. Example 7 H 2 SO 4  2H + + SO M M pH = - log[H + ] = -log( ) = -(-2.315) = 2.32

Homework Pg 239 #1, 2 Pg 242 #5, 7, 9, 10

Acid-Base Indicators Can determine if the solution is acidic, basic or natural using various indicators Litmus paper, bromothyomol blue, phenolphthalein are some examples. Depending on the indicator they change colour at varying pH levels. Need to use various indicators to solve the pH level

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