What is the pH of a 0.10 M solution of sodium cyanide (NaCN)? The pKa of hydrocyanic acid (HCN) is 9.21. A. 4.37 B. 5.43 C. 9.62 D. 11.1 I already know the answer but I was hoping someone could explain why?
The Correct Answer and Explanation is:
Correct Answer: D. 11.1
Explanation:
Sodium cyanide (NaCN) is a salt of a weak acid (HCN) and a strong base (NaOH). When NaCN dissolves in water, it dissociates completely into Na⁺ and CN⁻ ions. The Na⁺ ion does not affect the pH, but the CN⁻ ion can react with water in a process called hydrolysis:CN−+H2O⇌HCN+OH−\text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^-CN−+H2O⇌HCN+OH−
This reaction produces hydroxide ions (OH⁻), which makes the solution basic. To find the pH, we follow these steps:
Step 1: Find the Kb for CN⁻
We are given the pKa of HCN = 9.21. Use the relationship:pKa+pKb=14⇒pKb=14−9.21=4.79\text{pKa} + \text{pKb} = 14 \Rightarrow \text{pKb} = 14 – 9.21 = 4.79pKa+pKb=14⇒pKb=14−9.21=4.79
Now convert pKb to Kb:Kb=10−4.79≈1.62×10−5\text{Kb} = 10^{-4.79} \approx 1.62 \times 10^{-5}Kb=10−4.79≈1.62×10−5
Step 2: Use the ICE table and set up the expression:
Let CN⁻ = 0.10 M initially.Kb=[OH−]2[CN−]=x20.10\text{Kb} = \frac{[OH^-]^2}{[CN^-]} = \frac{x^2}{0.10}Kb=[CN−][OH−]2=0.10×21.62×10−5=x20.10⇒x2=1.62×10−6⇒x=[OH−]=1.62×10−6≈1.27×10−31.62 \times 10^{-5} = \frac{x^2}{0.10} \Rightarrow x^2 = 1.62 \times 10^{-6} \Rightarrow x = [OH^-] = \sqrt{1.62 \times 10^{-6}} \approx 1.27 \times 10^{-3}1.62×10−5=0.10×2⇒x2=1.62×10−6⇒x=[OH−]=1.62×10−6≈1.27×10−3
Step 3: Find the pOH and then the pH:
pOH=−log(1.27×10−3)≈2.90\text{pOH} = -\log(1.27 \times 10^{-3}) \approx 2.90pOH=−log(1.27×10−3)≈2.90pH=14−2.90=11.1\text{pH} = 14 – 2.90 = 11.1pH=14−2.90=11.1
Conclusion:
Since CN⁻ produces OH⁻ through hydrolysis, the solution becomes basic, and the pH rises above 7. With calculations complete, the pH of a 0.10 M NaCN solution is 11.1, which matches choice D.
