. Calculate the pH of a solution made by mixing 50.00 mL of 0.100 M NaCN with (a) 4.20 mL of 0.438 M HClO4 (b) 11.82 mL of 0.438 M HClO4 (c) What is the pH at the equivalence point with 0.438 M HClO4?
T he Correct Answer and Explanation is :
To calculate the pH of the solution after mixing NaCN (a weak base) and HClO₄ (a strong acid), we consider the acid-base reaction:
CN⁻ + H₃O⁺ → HCN + H₂O
Given:
- Volume of NaCN = 50.00 mL = 0.05000 L
- Molarity of NaCN = 0.100 M
- Moles of CN⁻ = 0.100 mol/L × 0.05000 L = 0.005000 mol
(a) Add 4.20 mL of 0.438 M HClO₄
- Volume HClO₄ = 4.20 mL = 0.00420 L
- Moles HClO₄ = 0.438 mol/L × 0.00420 L = 0.001840 mol
Since CN⁻ is in excess, this is a buffer solution (both CN⁻ and HCN present):
Moles CN⁻ after reaction = 0.005000 – 0.001840 = 0.003160 mol
Moles HCN formed = 0.001840 mol
Use the Henderson-Hasselbalch equation:
pKa of HCN = 9.21 (Ka = 6.2 × 10⁻¹⁰) pH=pKa+log([base][acid])=9.21+log(0.0031600.001840)\text{pH} = \text{pKa} + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right) = 9.21 + \log\left(\frac{0.003160}{0.001840}\right) pH≈9.21+log(1.717)≈9.21+0.235=9.45\text{pH} ≈ 9.21 + \log(1.717) ≈ 9.21 + 0.235 = \boxed{9.45}
(b) Add 11.82 mL of 0.438 M HClO₄
- Volume HClO₄ = 11.82 mL = 0.01182 L
- Moles HClO₄ = 0.438 mol/L × 0.01182 L = 0.005176 mol
All CN⁻ reacts and HClO₄ is in excess:
Moles excess HClO₄ = 0.005176 – 0.005000 = 0.000176 mol
Total volume = 50.00 mL + 11.82 mL = 61.82 mL = 0.06182 L
[H₃O⁺] = 0.000176 mol / 0.06182 L ≈ 2.847 × 10⁻³ M pH=−log(2.847×10−3)≈2.55\text{pH} = -\log(2.847 \times 10^{-3}) \approx \boxed{2.55}
(c) Equivalence point with 0.438 M HClO₄
Moles CN⁻ = 0.005000 mol
To reach equivalence, moles HClO₄ needed = 0.005000 mol
Volume HClO₄ = 0.005000 mol / 0.438 mol/L = 0.01142 L = 11.42 mL
At equivalence, all CN⁻ is converted to HCN (a weak acid). Now calculate pH from HCN dissociation: Ka=6.2×10−10,total volume=61.42 mL=0.06142 L\text{Ka} = 6.2 \times 10^{-10},\quad \text{total volume} = 61.42 \text{ mL} = 0.06142 \text{ L} [HCN]=0.0050000.06142=0.0814 M[\text{HCN}] = \frac{0.005000}{0.06142} = 0.0814\ \text{M}
Set up ICE table and solve: Ka=x20.0814=6.2×10−10⇒x=[H₃O⁺]=6.2×10−10×0.0814=7.1×10−6\text{Ka} = \frac{x^2}{0.0814} = 6.2 \times 10^{-10} \Rightarrow x = [\text{H₃O⁺}] = \sqrt{6.2 \times 10^{-10} \times 0.0814} = 7.1 \times 10^{-6} pH=−log(7.1×10−6)≈5.15\text{pH} = -\log(7.1 \times 10^{-6}) ≈ \boxed{5.15}
Summary:
- (a) pH = 9.45 (buffer solution)
- (b) pH = 2.55 (excess strong acid)
- (c) pH = 5.15 (weak acid at equivalence)
Explanation
This problem involves a weak base (NaCN, providing CN⁻) reacting with a strong acid (HClO₄). The key idea is to determine the pH based on the mole ratio of base to acid and understand the solution’s nature (buffer, excess acid, or equivalence).
In (a), less acid is added than base present. The strong acid partially neutralizes CN⁻, forming HCN. Since both the weak acid (HCN) and its conjugate base (CN⁻) coexist, the result is a buffer solution. We apply the Henderson-Hasselbalch equation to find pH, which reflects the buffer’s resistance to pH change.
In (b), more HClO₄ is added than the available CN⁻. All CN⁻ is neutralized, and the excess HClO₄ determines the solution’s pH. Since it’s a strong acid, the pH is directly calculated from the hydronium ion concentration using pH = -log[H₃O⁺]. The solution is now acidic, with no buffering capacity.
In (c), the amount of acid exactly neutralizes the base, forming HCN only. HCN, being a weak acid, partially ionizes, producing some H₃O⁺ ions. The pH is calculated using the Ka of HCN and solving the equilibrium expression from an ICE table. This results in a pH typical of a weak acid—lower than neutral but not as low as strong acid solutions.
Understanding these scenarios is fundamental in acid-base chemistry, particularly in titration curves and buffer systems.
