# The Equilibrium Constant Calculation for HCN Dissociation Reaction

## Equilibrium Constant Calculation for HCN Dissociation Reaction

In this problem, we are given the equilibrium constant Kc for the reaction HCN(aq) ⇌ H+(aq) + CN⁻(aq) as 4.9 × 10⁻¹⁰ at 25 °C. We need to calculate the equilibrium concentrations of all species at 25°C in a 0.150 M solution of aqueous HCN.

### Final answer:

The equilibrium concentrations are:
**[HCN]eq = 0.054 × 10¹ M**
**[H+]eq = 0.069 × 10¹ M**
**[CN⁻]eq = 0.069 × 10¹ M**

### Explanation:

Given the **chemical equilibrium** constant Kc for the reaction HCN(aq) ⇌ H+(aq) + CN⁻(aq) is 4.9 × 10⁻¹⁰ at 25°C, we can determine the equilibrium concentrations. Initially, [H+] = [CN⁻] = 0 as HCN has not dissociated. The concentration of HCN is 0.150 M.

Assuming x moles of HCN dissociate at equilibrium, we have [H+] = [CN⁻] = x and [HCN] = 0.150 - x. The equilibrium constant Kc = ([H+] [CN⁻])/[HCN]. Substituting these concentrations gives us (x)(x)/(0.150 - x) = 4.9 × 10^-10.

Since Kc is very small, we approximate that x is much smaller than 0.150, making (0.150 - x) ≈ 0.150. Solving for x gives us the equilibrium concentrations of all species.

## Do you understand the concept of chemical equilibrium and how to calculate equilibrium concentrations?

To better grasp the concept of chemical equilibrium and equilibrium concentrations, it is essential to practice various equilibrium problems and understand how to set up the necessary equations. The given example is just one of the many scenarios that students may encounter in chemistry courses. Remember to review the fundamentals of equilibrium constants and how they relate to reaction concentrations. Practice makes perfect!