What is the pH of a 1 x 10-7 M solution of HCl? Please note that you cannot simply add the contribution of H+ ions from the autoproteolysis of water and the contribution from the dissociation of HCl. You will need to use the quadratic equation to solve this problem and report the answer to 4 significant figures.

## Expert Answer

[H^{+}] from the acid = 10^{-7} M

Assume that the autoproteolysis of water produces ‘x’ M each of [H^{+}] and [OH^{–}].

So,

Net [H^{+}] = x + 10^{-7}

[OH^{–}] = x

Using relation:

Kw = [H^{+}]*[OH^{–}] = 10^{-14}

Putting values we get:

(x + 10^{-7})*x = 10^{-14}

So we get the following quadratic equation:

x^{2} + 10^{-7}*x – 10^{-14} = 0

Solving above equation we get:

x = 6.2*10^{-8}

So,

Net [H^{+}] = 10^{-7} + 6.2*10^{-8} = 1.6*10^{-7}.

Thus,

pH = 7-log(1.6) = **6.795**

Hope this helps !