Question

Calculate the [H⁺ ] in a 0.068 M solution of HCN,

Ka =6.2×10⁻¹⁰.
A) Determine
B) Evaluate
C) Compute
D) Analyze

Answer 1

The concentration of [H⁺] in the 0.068 M solution of HCN is approximately 2.4×10⁻⁴ M.

Step-by-step explanation:

To calculate the [H⁺] in a 0.068 M solution of HCN, we need to use the equilibrium expression for the dissociation of HCN, which is given by Ka = [H⁺][CN⁻]/[HCN].

First, let's assume that x is the concentration of [H⁺] in the solution.

Since there is a 1:1 stoichiometry between [H⁺] and [CN⁻], [CN⁻] will also be equal to x.

Therefore, we can write the expression as: Ka = x * x / (0.068 - x).

Substituting the values, we get: 6.2×10⁻¹⁰ = x² / (0.068 - x).

This is a quadratic equation. By solving this equation, we find that x ≈ 2.4×10⁻⁴ M.

Therefore, the concentration of [H⁺] in the solution is approximately 2.4×10⁻⁴ M.