Question

when 0.4 g of NaOH is dissolved in one litre of solution,the pH of solution is

Answer 1

Q. When 0.4 g of NaOH is dissolved in one litre of solution,the pH of solution ?

Ans :-
No. of moles of 0.4 g of NaOH
= 0.4/(23+16+1)
= 0.4/40
= 1/100
= 0.01 mole
Now molarity ( molar conentration )
of 0.01 mole of NaOH in 1 lit.
= mole / volume
= 0.01/1
= 0.01
= 10^(-2)

Now ,
$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: NaOH = ( OH^{ - })+ (Na^{ + }) \\ at \: t = 0 \: \: \: \: \: \: \: \: \: \: \: 10^{ - 2} \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0\\at \: t = t \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {10}^{ - 2} \: \: \: \: \: \: \: \: \: \: \: \: {10}^{ - 2} \\ \\ pOH \: = - log(OH^{ - } ) \\ = - log( {10}^{ - 2} ) \\ = 2 \\ \\ pOH + pH = 14 \\ pH = 14 - 2 = 12$