A chemist starts her experiments with 25.00 mL of 0.125 mol L-1 NaCN solution. The pKa of HCN = 9.20. a) What volume of 0.100 mol L-1 HCl will be required to fully react with the above salt? b) After the reaction in part a) takes places, will the pH of the solution be acidic, basic or neutral? c) What volume of 0.100 mol L-1 HCl is needed to adjust the pH of the originally provided solution to pH = 9.00?
The Correct Answer and Explanation is:
Given:
- Volume of NaCN solution = 25.00 mL = 0.02500 L
- Concentration of NaCN = 0.125 mol L⁻¹
- pKa of HCN = 9.20
- Concentration of HCl = 0.100 mol L⁻¹
Part (a): Volume of 0.100 mol L⁻¹ HCl required to fully react with NaCN
Reaction:
NaCN (aq) + HCl (aq) → HCN (aq) + NaCl (aq)
This is a 1:1 molar reaction between NaCN and HCl.
Moles of NaCN:
= 0.02500 L × 0.125 mol L⁻¹ = 0.003125 mol
Since the molar ratio is 1:1, 0.003125 mol of HCl is required.
Volume of HCl:
= moles / concentration = 0.003125 mol / 0.100 mol L⁻¹ = 0.03125 L
= 31.25 mL
Part (b): pH after full reaction
After all CN⁻ ions react with HCl, they convert to HCN, a weak acid. There will be no conjugate base left, so the solution contains only HCN.
Since HCN is a weak acid and no base is left, the pH will be acidic.
Answer: Acidic
Part (c): Volume of HCl to adjust the pH to 9.00
We are creating a buffer of HCN and CN⁻. Use the Henderson-Hasselbalch equation:pH=pKa+log([A−][HA])\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)pH=pKa+log([HA][A−])9.00=9.20+log([CN−][HCN])9.00 = 9.20 + \log\left(\frac{[\text{CN}^-]}{[\text{HCN}]}\right)9.00=9.20+log([HCN][CN−])−0.20=log([CN−][HCN])⇒[CN−][HCN]=10−0.20≈0.631-0.20 = \log\left(\frac{[\text{CN}^-]}{[\text{HCN}]}\right) \Rightarrow \frac{[\text{CN}^-]}{[\text{HCN}]} = 10^{-0.20} \approx 0.631−0.20=log([HCN][CN−])⇒[HCN][CN−]=10−0.20≈0.631
Let total moles of NaCN = 0.003125 mol
Let x be the moles of HCl added (which converts CN⁻ to HCN).
Then:
- CN⁻ remaining = 0.003125 – x
- HCN formed = x
0.003125−xx=0.631⇒0.003125−x=0.631x⇒0.003125=1.631x⇒x=0.001915 mol\frac{0.003125 – x}{x} = 0.631 \Rightarrow 0.003125 – x = 0.631x \Rightarrow 0.003125 = 1.631x \Rightarrow x = 0.001915 \text{ mol}x0.003125−x=0.631⇒0.003125−x=0.631x⇒0.003125=1.631x⇒x=0.001915 mol
Volume of HCl:
= 0.001915 mol / 0.100 mol L⁻¹ = 0.01915 L = 19.15 mL
Final Answers:
(a) 31.25 mL
(b) Acidic
(c) 19.15 mL
Explanation
This problem explores buffer chemistry and acid-base reactions. The initial solution contains sodium cyanide, a salt that dissociates fully into sodium and cyanide ions. The cyanide ion is the conjugate base of hydrocyanic acid (HCN), a weak acid with a pKa of 9.20.
In part (a), adding HCl to NaCN leads to a direct neutralization reaction. HCl donates a proton to CN⁻, forming HCN. Since the stoichiometry is 1:1, the number of moles of HCl needed to fully react with CN⁻ equals the moles of NaCN present, which is calculated using the volume and concentration. This results in a requirement of 31.25 mL of HCl solution.
In part (b), after the reaction completes, all CN⁻ has been converted to HCN, leaving no base to neutralize additional acid. Since HCN is a weak acid and is the only species in solution contributing to the pH, the solution will be acidic.
Part (c) introduces a classic buffer situation. To reach a pH of 9.00, the chemist must partially convert CN⁻ to HCN. The Henderson-Hasselbalch equation allows us to relate the pH to the ratio of base to acid. Using the equation, we calculate the number of moles of HCl needed to reach the desired ratio and then use the molarity of HCl to find the volume. This comes out to 19.15 mL, which is less than the full neutralization volume, as expected in a buffer system.
The solution emphasizes stoichiometry, equilibrium, and buffer theory, which are essential concepts in analytical and preparative chemistry.
