AP Chemistry · Lesson 1

pH, strong acids and weak acids

Know Know it

pH=−log⁡[H3O+]\text{pH} = -\log[\text{H}_{3}\text{O}^{+}] and pOH=−log⁡[OH−]\text{pOH} = -\log[\text{OH}^{-}]. Water itself ionizes slightly, and Kw=[H3O+][OH−]=1.0×10−14K_w = [\text{H}_{3}\text{O}^{+}][\text{OH}^{-}] = 1.0 \times 10^{-14} at 25 °C, so pH+pOH=14.00\text{pH} + \text{pOH} = 14.00 at that temperature. Each pH unit is a factor of 10 in [H3O+][\text{H}_{3}\text{O}^{+}].

A solution is neutral when [H3O+]=[OH−][\text{H}_{3}\text{O}^{+}] = [\text{OH}^{-}]. That is pH 7.00 only at 25 °C: KwK_w grows with temperature, so neutral water is below pH 7 when it is warmer.

Strong acids (HCl, HBr, HI, HNO3\text{HNO}_3, HClO4\text{HClO}_4, and the first step of H2SO4\text{H}_2\text{SO}_4) ionize completely, so a monoprotic strong acid gives [H3O+][\text{H}_{3}\text{O}^{+}] equal to its concentration. Strong bases such as NaOH and Ba(OH)2\text{Ba(OH)}_2 dissociate completely; count every hydroxide in the formula.

Significant figures with logs: the number of decimal places in a pH equals the number of significant figures in the concentration. A concentration of 2.0×10−32.0 \times 10^{-3} M gives pH 2.70, not 2.7.

Calculator On the exam

On the exam you may use a scientific or graphing calculator on both sections (a four-function calculator is allowed but is a poor choice). You also get a reference booklet of chemistry equations and constants. Use the log and 10x10^x keys for every pH step, and keep two decimal places in pH for two significant figures in concentration.

Worked example Easy

What is the pH of a 0.0050 M solution of barium hydroxide, Ba(OH)2\text{Ba(OH)}_2, at 25 °C?

  1. 11.70
  2. 2.00
  3. 12.00Answer
  4. 2.30

How to solve it

  1. Ba(OH)2\text{Ba(OH)}_2 is a strong base and gives two OH−\text{OH}^- per unit: [OH−]=2(0.0050)=0.010[\text{OH}^{-}] = 2(0.0050) = 0.010 M.
  2. pOH=−log⁡(0.010)=2.00\text{pOH} = -\log(0.010) = 2.00.
  3. pH=14.00−2.00=12.00\text{pH} = 14.00 - 2.00 = 12.00 at 25 °C.

Why each choice is right or wrong

  • A. Incorrect. A pH of 11.70 uses [OH−]=0.0050[\text{OH}^{-}] = 0.0050 M, forgetting that each Ba(OH)2\text{Ba(OH)}_2 releases two hydroxide ions.
  • B. Incorrect. A pH of 2.00 is the pOH; the pH is 14.00 minus that value at 25 °C.
  • C. Correct. Each formula unit gives two hydroxide ions, so [OH−]=0.010[\text{OH}^{-}] = 0.010 M, the pOH is 2.00 and the pH is 14.00 - 2.00 = 12.00.
  • D. Incorrect. A pH of 2.30 is the pOH you get from 0.0050 M hydroxide, and it is a pOH, not a pH.

Trap Neutral means equal, not pH 7

A solution is neutral when hydronium and hydroxide are equal. At 25 °C that happens at pH 7.00, but at other temperatures KwK_w is different and so is the neutral pH. And a weak acid is not "fully ionized": never set [H3O+][\text{H}_{3}\text{O}^{+}] equal to a weak acid’s concentration.

Explain Explain it

Task verbs: calculate, compare

A weak acid HA only partly ionizes: HA+H2O⇌H3O++A−\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-, with Ka=[H3O+][A−][HA]K_a = \frac{[\text{H}_{3}\text{O}^{+}][\text{A}^-]}{[\text{HA}]}. Starting from concentration C, the ICE table gives Ka=x2/(C−x)K_a = x^2/(C - x). When x is small next to C (under about 5 percent), x≈KaCx \approx \sqrt{K_a C}.

A weak base B works the same way with KbK_b and gives [OH−]≈KbC[\text{OH}^{-}] \approx \sqrt{K_b C}. For a conjugate pair, KaKb=KwK_a K_b = K_w: the weaker the acid, the stronger its conjugate base. So the salt of a weak acid, like sodium acetate, makes a basic solution.

Worked example Medium

A weak acid HZ has Ka=6.3×10−5K_a = 6.3 \times 10^{-5}. What is the pH of a 0.20 M solution of HZ at 25 °C?

  1. 0.70
  2. 2.45Answer
  3. 4.20
  4. 4.90

How to solve it

  1. ICE: Ka=x2/(0.20−x)≈x2/0.20K_a = x^2/(0.20 - x) \approx x^2/0.20.
  2. x=(6.3×10−5)(0.20)=3.5×10−3x = \sqrt{(6.3 \times 10^{-5})(0.20)} = 3.5 \times 10^{-3} M, which is under 5 percent of 0.20 M, so the shortcut holds.
  3. pH=−log⁡(3.5×10−3)=2.45\text{pH} = -\log(3.5 \times 10^{-3}) = 2.45.

Why each choice is right or wrong

  • A. Incorrect. A pH of 0.70 treats HZ as a strong acid, setting [H3O+][\text{H}_{3}\text{O}^{+}] equal to 0.20 M, but HZ only partly ionizes.
  • B. Correct. [H3O+]≈(6.3×10−5)(0.20)=3.5×10−3[\text{H}_{3}\text{O}^{+}] \approx \sqrt{(6.3 \times 10^{-5})(0.20)} = 3.5 \times 10^{-3} M, so the pH is 2.45.
  • C. Incorrect. A pH of 4.20 is the pKa\text{p}K_a of HZ, which equals the pH only when HZ and Z−\text{Z}^- are present in equal amounts.
  • D. Incorrect. A pH of 4.90 takes the log of KaCK_a C without the square root, but x2=KaCx^2 = K_a C, so x is the square root.

Calculate Your turn

A 0.25 M solution of a weak base B has Kb=6.4×10−5K_b = 6.4 \times 10^{-5} at 25 °C. Calculate the hydroxide concentration and the pH of the solution.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Finds the hydroxide concentration from KbK_b, 4.0×10−34.0 \times 10^{-3} M
  • Converts to pH, about 11.60

Model answer

B reacts with water: B+H2O⇌BH++OH−\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-, so Kb=x2/(0.25−x)≈x2/0.25K_b = x^2/(0.25 - x) \approx x^2/0.25. Then [OH−]=x=(6.4×10−5)(0.25)=4.0×10−3[\text{OH}^{-}] = x = \sqrt{(6.4 \times 10^{-5})(0.25)} = 4.0 \times 10^{-3} M, which is under 5 percent of 0.25 M. The pOH is −log⁡(4.0×10−3)=2.40-\log(4.0 \times 10^{-3}) = 2.40, so the pH is 14.00 - 2.40 = 11.60.

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