What is the hydronium ion concentration in a solution prepared by mixing 150.0 mL of 0.10 M HCN with 150.0 mL of 0.050 M NaCN? Assume that the volumes of the solutions are additive and that K_a = 4.9 × 10^-¹⁰ for HCN. A) 2.4 × 10^-¹⁰ M B) 4.9 × 10^-10 M C) 9.8 × 10^-¹⁰ M D) 7.0 × 10^-6 M

Question

Quizplus · Accepted Answer

First, we need to write out the chemical equation for the reaction between HCN and NaCN:

HCN + NaCN <-> H+ + CN- + Na+

Since Na+ is a spectator ion and does not affect the acidity or basicity of the solution, we can ignore it and focus on the HCN and CN- ions.

Using the equilibrium constant expression for this reaction, we get:

K = [H+][CN-]/[HCN]

We are given the initial concentrations of HCN and NaCN, and we can assume that the volumes of the solutions are additive. Thus, the total volume of the solution is 300.0 mL or 0.300 L.

To find the final concentration of CN-, we need to use the stoichiometry of the reaction. Since HCN and NaCN react in a 1:1 ratio, the final concentration of CN- is equal to the initial concentration of NaCN:

[CN-] = 0.050 M

To find the final concentration of HCN, we need to use the equilibrium constant expression and solve for [H+]:

K = [H+][CN-]/[HCN]
4.9 × 10^-10 = [H+](0.050 M)/0.10 M
[H+] = 9.8 × 10^-10 M

Finally, we can use the definition of pH to find the hydronium ion concentration:

pH = -log[H+]
[H+] = 10^-pH
[H+] = 9.8 × 10^-10 M

Therefore, the hydronium ion concentration in the solution is 9.8 × 10^-10 M, which corresponds to choice C.

What Is the Hydronium Ion Concentration in a Solution Prepared