Calculate the pH of a 1.33 mol/L solution of HCN is the Ka value of HCN is 6.2x10^-12

Set up ionization.

Write Ka expression.
Set up ICE table.
Substitute and solve for (H^+), then convert that to pH.
I'll be glad to help you through any of it if you post your work but I am not interested in typing out the problem so you can check yuor answers.

To calculate the pH of a solution of HCN (hydrogen cyanide) with a given concentration and Ka value, we can use the equation for the acid dissociation constant.

The equation for Ka is defined as follows:

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

First, we need to determine the concentration of H+ ions. Since HCN is a weak acid, we assume that it dissociates only slightly, and thus, [HCN] will be approximately equal to the initial concentration.

Given that the concentration of HCN is 1.33 mol/L, we can assume [HCN] = 1.33 mol/L.

Note: The brackets [] represent the concentration of the substance.

Now, we need to determine the concentration of [CN-] and [H+] ions. Since HCN dissociates to produce equal amounts of CN- and H+ ions, we can assume that [CN-] = [H+].

Let's denote this concentration as x.

Now, we can rewrite the equation for Ka as follows:

Ka = (x)(x) / (1.33 - x)

Since the concentration of H+ ions and CN- ions is x, we substitute that into the equation.

6.2x10^-12 = (x)(x) / (1.33 - x)

Solving this quadratic equation will give us the value of x, which represents the concentration of H+ ions and CN- ions.

After solving the equation, we find that the value of x is approximately 2.96x10^-6 mol/L.

Now, we can calculate the pH using the equation:

pH = -log[H+]

Since [H+] = x, substitute the value of x into the equation:

pH = -log(2.96x10^-6) ≈ 5.53

Therefore, the pH of a 1.33 mol/L solution of HCN with a Ka value of 6.2x10^-12 is approximately 5.53.