What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 M HCN with 50.00 mL of 0.030 M NaCN? Assume that the volumes of the solutions are additive and that
K_a = 4.9 × 10^-10 for HCN.

Question

Quizplus · Accepted Answer

First, we must write out the equation for the dissociation of HCN:

HCN + H2O ⇌ H3O+ + CN-

We can then write out the equilibrium expression using the acid dissociation constant, Ka:

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

Substituting the given values:

Ka = (x)(0.030)/(0.10-x)

Since the initial concentrations of HCN and CN- are both 0.10 M and 0.030 M, respectively, we can assume that x (the concentration of H3O+) is much less than 0.10 M, and thus 0.10 - x ≈ 0.10.

Therefore,

Ka = (x)(0.030)/(0.10) = 4.9 × 10^-10

Solving for x:

x = (4.9 × 10^-10)(0.10)/0.030 = 1.6 × 10^-9

Thus, the hydronium ion concentration in the solution is 1.6 × 10^-9 M.

The best choice is (C) 1.6 × 10^-9 M.

What Is the Hydronium Ion Concentration in a Solution Prepared