Question

2 )5 mL 0.1 M NH 4OH and 250 mL 0.1 M NH4Cl are
mixed together. Calculate the pH of the solution.
Kb = 1.8 × 10-5.
3) How many moles of CH3COOH should be added to 0.435 moles of CH3COONa so that pH of the solution is 4.56? Given that pKa 4.7.
​

Answer 1

Number of gram equivalents(eq) = moles x acidity of a base (or bascity of an acid.)

Or

= Molarity x Volume in litres x acidity of a base (or bascity of an acid)

So,

Eq(NaOH) = 50/1000 x 0.1 x 1 = 0.005 eq

Eq(CH3COOH) = 20/1000 x0.25 x1 = 0.005 eq

That means acid = base. Hence a salt would be formed in the solution with total volume = 50 + 20 = 70 ml

The pH of a strong base weak acid salt solution =

7 + 0.5(pKa +logC) where C is concentration of salt

concentration of salt = eq/volume =1000 x 0.005/70

= 0.0714

So

pH = 7 + 0.5(4.74 +log0.0714)

pH = 8.8

For such questions always first determine what will exist after the mixture has reacted among itself. Once you know what is present, you can apply the formula.

Hope this helps.