Question

Chester makes 1L of a 0.01M HCl solution and measure the pH. Burt then adds 5.80 g of sodium chloride
to Chester’s solution and measures the pH. Calculate the difference between the two pH values
obtained.

Answer 1

pH = -LOG[H+]

This means that the pH is the -log of the [H+] which is the concentration of acid.

Answer 2

because HCl dissociates completely, we know that we'll have 0.01M of H+

Answer 3

@MASTERPARKOUR

Answer 4

I understand that, but how do I calculate the change in pH after the addition of NaCl?

Answer 5

@aaronq @sweetburger

Answer 6

There won't be any noticeable change in pH because the dissociation constant for HCl is really high, 1.3*10^6.

Here the chloride ion would behave as a base, thus the increase in \([Cl^-]\) would cause the reformation of HCl, decreasing \([H^+]\) available, in the reaction:

\(Cl^-+H^+\rightleftharpoons HCl\)

The K for this reaction is the reciprocal of the dissociation constant for HCl.

\(K=\dfrac{1}{1.3*10^6}=7.7*10^{-7}\)

\(\sf K=\dfrac{[HCl]}{[H^+][Cl^-]}=\dfrac{[HCl]}{(0.01-[HCl])((0.01+0.1)-[HCl])}\)

(Note: I skipped showing the I.C.E table for the result above)

\(\sf [HCl]\approx8.5*10^{-10}M\)

The amount of \([H^+]\) is 8 orders of magnitude lower, so it's very insignificant.

Answer 7

The amount of \([H^+]\) taken up** i meant.