Question

How many grams of NaCN would you need to dissolve in enough water to make exactly 250 mL of solution with a pH of 10.00?

(Ka, HCN = 4.9 X 10^-10)

Answer 1

pH + pOH = 14

pOH = 14 – 10 = 4

[OH^-] = $10^{-4} \;M$

NaCN + H₂O <=> HCN + OH^-

To calculate the initial molarity of NaCN:

Let the concentration of NaCN be x such that

Equilibrium:

[NaCN] = (x – 10^{-4}) \; M \\ [HCN] = 10^{-4} \;M \\ [OH^-] = 10^{-4} \;M \\ K_b = \frac{[HCN][OH^-]}{[NaCN]} \\ K_b = \frac{K_w}{K_a} \\ K_b = \frac{10^{-14}}{4.9 \times 10^{-10}} = 2 \times 10^{-5} \\ 2 \times 10^{-5} = \frac{10^{-4} \times 10^{-4}}{(x – 10^{-4})} \\ x = 6 \times 10^{-4} \\ [NaCN] = 6 \times 10^{-4} \;M \\ n(NaCN) = 6 \times 10^{-4} \;M \times 0.250 \;L = 1.5 \times 10^{-4} \;mol \\ M(NaCN) = 49.01 \;g/mol \\ m(NaCN) = 49.01 \times 1.5 \times 10^{-4} = 7.35 \times 10^{-3} \;g

Answer: $7.35 \times 10^{-3} \;g$