30 mL 0.2(N) `H_2SO_4` solution is mixed with 20 mL 0.3(N) `H_2SO_4` . What will be the pH of the mixed solution?

Explanation

0.2g NaOH`=(0.2)/(40)=5xx10^(-3)mol` NaOH
Amount of NaOH present per litre of NaOH solution
`=5xx10^(-3)mol`
Hence, amount of NaOH present in 250 mL of this
solution`=(5xx10^(-3))/(1000)xx250=1.25xx10^(-3)mol`
=1.25 million mol
12.5 mL 0.1(N) `H_2SO_4-=12.5mL 0.05(M)H_2SO_4`
12.5 mL 0.1(N) `H_2SO_4-=0.625` million mol `H_2SO_4`
NaOH reacts with `H_2SO_4` as follows :
`2NaOH(aq)+H_2SO_4(aq)toNa_2SO_4(aq)+2H_2O(l)`
Hence, 1 mol of `H_2SO_4` reacts completely with 2 mol of Na OH to produce salt and water. Therefore, 0.625 millimol `H_2SO_4` reacts completely with
`2xx0.625=1.25` millimol NaOH and there is no residual `H_2SO_4` or NaOH in the solution. Again since `Na_2SO_4` is is a salt of strong acid and strong base, it is not hydrolysed in the solution i.e., the mixed solution is neutral in nature and obviously pH of the mixed solution is 7