Question 18.47: Write a balanced chemical equation and equilibrium expressio......
Write a balanced chemical equation and equilibrium expression for the dissociation of chloroacetic acid and convert \text{p}K_\text{a}\text{ to }K_\text{a}.
Step-by-Step
K_\text{a} = 10^{-\text{p}K} = 10^{-2.87} = 1.34896 \times 10^{-3}\text{ (unrounded)}
\begin{matrix} \text{ClCH}_2\text{COOH}(aq)& + \text{H}_2\text{O}(l)& \leftrightarrows &\text{H}_3\text{O}^+(aq) +& \text{ClCH}_2\text{COO}^-(aq) \\ 1.05 – \text{x} &&& \text{x}& \text{x} \end{matrix}
K_\text{a}=1.34896\times 10^{-3}=\frac{(\text{H}_3\text{O}^+)(\text{ClCH}_2\text{COO}^-)}{(\text{ClCH}_2\text{COOH})} \\ K_\text{a}=1.34896\times 10^{-3}=\frac{(\text{x})(\text{x})}{(1.05-\text{x})} \quad \text{Assume x is small compared to 1.05.} \\ K_\text{a}=1.34896\times 10^{-3}=\frac{(\text{x})(\text{x})}{(1.05)} \\ \text{x}=0.037635\text{ (unrounded)}
Check assumption: (0.037635 / 1.05) x 100% = 4%. The assumption is good.
[\text{H}_3\text{O}^+] = [\text{ClCH}_2\text{COO}^-] = 0.038 \ M \\ [\text{ClCH}_2\text{COOH}] = 1.05 – 0.037635 = 1.012365 = 1.01 \ M \\ \text{pH} = -\log [\text{H}_3\text{O}^+] = -\log (0.037635) = 1.4244 = 1.42
Related Answered Questions
Question: 18.p.10
Verified Answer:
a) Since the base is strong and the acid is weak, ...
Question: 18.34
Verified Answer:
\text{a) }\sout{\text{Na}}^+(\sout{aq})+\te...
Question: 18.p.9
Verified Answer:
\begin{matrix} & \text{ClO}^-(aq)+ &...
Question: 18.p.8
Verified Answer:
\text{C}_5\text{H}_5\text{N}(aq) + \text{H}...
Question: 18.p.7
Verified Answer:
\begin{matrix} & \text{HOCN}(aq)+ &...
Question: 18.p.6
Verified Answer:
\begin{matrix} & \text{NH}_4^+(aq)+ &am...
Question: 18.99
Verified Answer:
M\text{ H}_2\text{O}_2=\left\lgroup\frac{1....
Question: 18.p.2
Verified Answer:
Calculating [\text{H}_3\text{O}^+] : \\ [\t...
Question: 18.p.3
Verified Answer:
pH + pOH = 14.00
pOH = 14.00 - 9.52 = 4.48
...
Question: 18.p.4
Verified Answer:
a) \text{CH}_3\text{COOH has one more H}^+\...