In order to create a buffer, 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 12.0 mL of 0.300 M NaH? K_a = 4.9 × 10⁻¹⁰.

• A. 6.82
• B. 10.01
• C. 8.52
• D. 10.17

Answer: (D)

Verified step by step guidance

1. Calculate the moles of sodium cyanide (NaCN) initially present. Use the molar mass of NaCN to convert grams to moles: \(\text{moles NaCN} = \frac{7.510\ \text{g}}{\text{molar mass of NaCN}}\).
2. Calculate the moles of hydrocyanic acid (HCN) initially present using its concentration and volume: \(\text{moles HCN} = 0.250\ \text{M} \times 0.1000\ \text{L}\).
3. Calculate the moles of sodium hydride (NaH) added: \(\text{moles NaH} = 0.300\ \text{M} \times 0.0120\ \text{L}\). Since NaH is a strong base, it will react with HCN to form CN$^-$ and water, so subtract these moles from HCN and add them to CN$^-$.
4. Determine the new moles of HCN and CN$^-$ after the reaction with NaH: \(\text{moles HCN}_{new} = \text{moles HCN}_{initial} - \text{moles NaH}\) and \(\text{moles CN}^-_{new} = \text{moles NaCN}_{initial} + \text{moles NaH}\).
5. Use the Henderson-Hasselbalch equation to calculate the pH of the buffer: \(\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\mathrm{CN}^-]}{[\mathrm{HCN}]} \right)\), where \(\mathrm{p}K_a = -\log K_a\). Use the new mole values and total volume (initial volume + volume of NaH added) to find concentrations.