Question

One litre solution contains 1m hocl [ka = 10-8] and 1 m naoh what is the ph of the solution

Answer 1

The pH of the solution is 4.5

Explanation:

HOCl(aq) + H2O (l)--> H3O+ (aq) + ClO- (aq).  

Initial concentrations before dissociation are

0.1 M HOCl, 0 M H3O+ and 0 M ClO- .

After dissociation, the concentration of HOCl drops by some amount "x" (concentration = 0.1-x) and the concentrations of H3O+ and ClO-  each becomes (0 +x) amount.

Substituting into the equation.  

Ka =  10^-8 = x^2/(0.1-x)  

Now assuming that x is small relative to 0.1, and therefore 0.1-x is to a good approximation 0.1.  

That means 10^-8 = x^2/0.1, So x^2 = 10-9.  

Therefore x = 3.16 x 10^-5 M.

X is the H3O+ concentration,

pH = -log(3.16 x 10^-5) = 4.5

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