What is the hydronium ion concentration in a solution prepared by mixing 50.00 mL of 0.10 mol L^-1 HCN with 50.00 mL of 0.030 mol L^-1 NaCN? Assume that the volumes of the solutions are additive and that K_a = 4.9 × 10^-10for HCN. A)1.5 × 10^-10 mol L^-1 B)4.9 × 10^-10 mol L^-1 C)1.6 × 10^-9 mol L^-1 D)7.0 × 10^-6 mol L^-1

Question

Quizplus · Accepted Answer

Step 1: Calculate the moles of HCN and NaCN in each solution.

Moles of HCN = 0.10 mol/L x 0.050 L = 0.005 mol
Moles of NaCN = 0.030 mol/L x 0.050 L = 0.0015 mol

Step 2: Calculate the total moles of cyanide ions in the solution.

Moles of CN- = 0.0015 mol + 0.005 mol = 0.0065 mol

Step 3: Calculate the concentration of hydronium ions using the acid dissociation constant (Ka) for HCN.

Ka for HCN = 4.9 x 10^-10

HCN + H2O <-> H3O+ + CN-

Initial concentration of HCN = 0.005 mol/0.100 L = 0.05 M
Initial concentration of CN- = 0.0065 mol/0.100 L = 0.065 M

Let x be the concentration of H3O+.

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

4.9 x 10^-10 = x(0.065)/(0.05-x)

Assuming x<<0.05, we can simplify the equation as follows:

4.9 x 10^-10 = x(0.065)/0.05

x = 1.6 x 10^-9 M

Therefore, the hydronium ion concentration in the solution is 1.6 x 10^-9 mol/L or 1.6 x 10^-9 M.

The correct answer is C) 1.6 × 10^-9 mol/L.

What Is the Hydronium Ion Concentration in a Solution Prepared