Question 15.7: Finding the pH of a Weak Acid Solution in Cases Where the x ...
Finding the pH of a Weak Acid Solution in Cases Where the x is small Approximation Does Not Work
Find the pH of a 0.100 M HClO_{2} solution.
| 1. Write the balanced equation for the ionization of the acid and use it as a guide to prepare an ICE table showing the given concentration of the weak acid as its initial concentration.
(Note that the H_{3}O^{+} concentration is listed as approximately zero. Although a little H_{3}O^{+} is present from the autoionization of water, this amount is negligibly small compared to the amount of H_{3}O^{+} from the acid.) |
HClO_{2}(aq) + H_{2}O(l) \xrightleftharpoons[]{} H_{3}O^{+}(aq) + ClO_{2}^{-}(aq)
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| 2. Represent the change in [ H_{3}O^{+} ] with the variable x. Define the changes in the concentrations of the other reac-tants and products in terms of x. |
HClO_{2}(aq) + H_{2}O(l) \xrightleftharpoons[]{} H_{3}O^{+}(aq) + ClO_{2}^{-}(aq)
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| 3. Sum each column to determine the equilibrium concentrations in terms of the initial concentrations and the variable x. | HClO_{2}(aq) + H_{2}O(l) \xrightleftharpoons[]{} H_{3}O^{+}(aq) + ClO_{2}^{-}(aq)
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| 4. Substitute the expressions for the equilibrium concentrations (from step 3) into the expression for the acid ionization con stant (K_{a}). Make the x is small approximation and substitute the value of the acid ionization constant (from Table 15.5) into the K_{a} expression. Solve for x.
Check to see if the x is small approximation is valid by calculating the ratio of x to the number it was subtracted from in the approximation. The ratio should be less than 0.05 (or 5%). |
K_{a} =\frac{ [H_{3}O^{+}][ClO_{2}^{-}]}{[HNO_{2}]} =\frac{ x_{2}}{0.100 – x}(x is small) 0.011 =\frac{ x^{2}}{0.100} \sqrt{0.011} = \sqrt{\frac{x^{2}}{0.100}} x = \sqrt{(0.100)(0.011)} = 0.033 \frac{0.033}{0.100}× 100% = 33% Therefore, the x is small approximation is not valid. |
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| 4a.If the x is small approximation is not valid, solve the quadratic equation explicitly or use the method of successive approxima tions to find x. In this case, the quadratic equation is | 0.011 =\frac{ x^{2}}{0.100 – x} 0.011(0.100 – x) = x² 0.0011 – 0.011x = x² x² + 0.011x – 0.0011 = 0 x =\frac{ -b ±\sqrt{ b^{2} – 4ac}}{2a} = \frac{-(0.011) ± \sqrt{(0.011)^{2} – 4(1)(-0.0011)}}{2(1)} = \frac{-0.011 ± 0.06\underline{7}2}{2} x = -0.039 or x = 0.028 |
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| Since x represents the concentration of H_{3}O^{+}, and since concentrations cannot be negative, reject the negative root. x = 0.028 |
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| 5. Determine the H_{3}O^{+} concentration from the calculated value of x and calculate the pH (if necessary). | [ H_{3}O^{+} ] = 0.028 M pH = -log[ H_{3}O^{+} ] = -log0.028 = 1.55 |
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| 6. Check your answer by substituting the calculated equilibrium values into the acid ionization expression. The calculated value of K_{a} should match the given value of K_{a}. Note that rounding errors could cause a difference in the least signifi cant digit when comparing values of K_{a}. | K_{a} = \frac{[H_{3}O^{+}][ClO_{2}^{-}]}{[HClO_{2}]} = \frac{0.028^{2}}{0.100 – 0.028} = 0.011 Since the calculated value of K_{a} matches the given value, the answer is valid. |
| TABLE 15.5 Acid Ionization Constants (K_{a}) for Some Monoprotic Weak Acids at 25 °C | |||||
| Acid | Formula | Structural Formula | Ionization Reaction | K_{a} | pK*_{a} |
| Chlorous acid | HCIO_{2} | H—O—Cl=O | HClO_{2}(aq)+H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+ClO_{2}^{-}9aq) | 1.1×10^{-2} | 1.96 |
| Nitrous acid | HNO_{2} | H—O—N=O | HNO_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + NO_{2}^{–}(aq) | 4.6×10^{-4} | 3.34 |
| Hydrouoric acid | HF | H—F | HF(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + F^{–}(aq) | 3.5×10^{-4} | 3.46 |
| Formic acid | HCHO_{2} | H—O—\overset{O}{\underset{C}{\parallel } } —H | HCHO_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O+(aq) + CHO_{2}^{–}(aq) | 1.8×10^{-4} | 3.74 |
| Benzoic acid | HC_{7}H_{5}O_{2} |  |
HC_{7}H_{5}O_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + C_{7}H_{5}O_{2}^{–}(aq) | 6.5×10^{-5} | 4.19 |
| Acetic acid | HC_{2}H_{3}O_{2} | H—O\overset{O}{\underset{C}{\parallel } } —CH_{3} | HC_{2}H_{3}O_{2}(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + C_{2}H_{3}O_{2}^{–}(aq) | 1.8×10^{-5} | 4.74 |
| Hypochlorous acid | HCIO_{2} | H—O—CI | HClO(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq) + ClO^{–}(aq) | 2.9×10^{-8} | 7.54 |
| Hydrocyanic acid | HCN | H—C≡N | HCN(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+CN^{–}(aq) | 4.9×10^{-10} | 9.31 |
| Phenol | HC_{6}H_{5}O |  |
HC_{6}H_{5}O(aq)+ H_{2}O(l)\xrightleftharpoons[]{}H_{3}O^{+}(aq)+C_{6}H_{5}O^{–}(aq) | 1.3×10^{-10} | 9.89 |
| *pK_{a} = –logK_{a} (See the discussion at the end of Section 15.5) | |||||
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