How much solid NaCN must be added to 1.0 L of a 0.5 M HCN solution to produce a solution with pH 7.0? K_a = 6.2 × 10^-10 for HCN.

Question

Quizplus · Accepted Answer

Step 1: Write the equation for the dissociation of HCN in water:

HCN + H2O ⇌ CN- + H3O+

Step 2: Write the expression for the acid dissociation constant (Ka) of HCN:

Ka = [CN-][H3O+]/[HCN]

Step 3: Use the given value of Ka (6.2 × 10-10) to calculate the concentration of CN- and H3O+ in the 0.5 M HCN solution at equilibrium:

6.2 × 10-10 = [CN-][H3O+]/0.5
[CN-][H3O+] = 3.1 × 10-10

Assuming that the initial concentration of H3O+ is negligible compared to the concentration at equilibrium, the concentration of CN- is approximately equal to 3.1 × 10-10 M.

Step 4: Calculate the concentration of Na+ that will be produced when solid NaCN dissolves in water:

NaCN(s) ⇌ Na+(aq) + CN-(aq)

Assuming that NaCN is completely dissociated, the concentration of Na+ will be equal to the concentration of CN- added.

Step 5: Use the concentration of Na+ to calculate the amount of NaCN needed to produce a solution with pH 7.0:

pH = 7.0
[H3O+] = 10-pH = 1.0 × 10-7 M
Ka = [CN-][H3O+]/[HCN] = 6.2 × 10-10

Assuming that the concentration of HCN remains approximately equal to 0.5 M, the concentration of CN- must be increased by a factor of 10-7/6.2 × 10-10 = 1613. This means that the concentration of Na+ must also be increased by a factor of 1613.

1.0 L of 0.5 M HCN solution contains 0.5 mol of HCN. To increase the concentration of Na+ by a factor of 1613, we need to add 1613 × 0.5 mol of NaCN:

mol of NaCN = 1613 × mol of Na+ = 1613 × [CN-] = 1613 × 3.1 × 10-10 M × 1.0 L = 0.00050 mol

mass of NaCN = 0.00050 mol × 49 g/mol = 0.0245 g ≈ 0.15 g

Therefore, the best answer is B.

How Much Solid NaCN Must Be Added to 1