Question

Calculate the pH of an aqueous solution containing 1.0 x 10^-2 M HCL, 1.0 x 10^-2 M H2SO4 (Ka1 = Ka2 = 1.2 x 10^-2 M HCN (Ka = 6.2 x10^-10).

Answer 1

This is not nearly as easy as it looks but it has solution.

The simple way to do it is to add H+ from 0.01M HCl, (It's a strong acid and it dissociates completely in solution so concentration of HCl = H+), 0.01 M H₂SO₄(first H+ ONLY) and ignore the H+ from k2 for H₂SO₄ and H+ from HCN (because is a weak acid).

You can go through the math but the second H from H₂SO₄ contributes only 0.0046 but compared to 0.02 M from above probably should be added in. If you want to do that use:

r: HSO₄^- ---------> SO₄^2- + H⁺

i: 0.01 0 0.02

e: 0.01-x x 0.02+x

Ka2 = (H+)(SO4)/(HSO4). For (H+) substitute the values and solve for x:

0.0102 = x(0.02+x) / (0.01-x)

0.0102(0.01-x) = x² + 0.02x

0.000102 - 0.0102x = x² + 0.02x

x² + 0.0098x - 0.000102 = 0

Using the quadratic formula for x, the two possible values are:

x₁ = 0.0063 M; x₂ = -0.0063 M

I got 0.0063M for Ka2 contribution from H₂SO₄. For the HCN, it's a weak acid, and it's contribution to pH can be neglected. This is because Ka is really small and the value of H⁺ would be very small too, so it's contribution can be neglected.

Now that we know the value of x, we can know the value of [H+] to get pH:

pH = -log(0.02+0.0063)

pH = 1.58

Hope this helps