100 ml of 0.1 N NaOH is mixed with 100 ml of 0.1 N H_{2}SO_{4}.The pH of the resultant solution is??

Normality =0.1 N

molarity=normality= 0.1M

moles of NaOH = 0.1$\times $0.1 = 0.01mol [using formula mole = C (concentration in M) $\times $V in L]

H

_{2}SO

_{4}volume = 100ml or 0.1L

Normality= 0.1 N

number of electron change (n) for H

_{2}SO

_{4}= 2

Thus molarity = 0.05 M ( since N = M $\times $n)

Moles of H

_{2}SO

_{4}= 0.1$\times $0.1 = 0.005 mol

0.01 mole of NaOH exactly neutralizes 0.005 mole of H

_{2}SO

_{4}according to the reaction

NaOH + H

_{2}SO

_{4}$\to $Na

_{2}SO

_{4}+ H

_{2}O

Thus neither NaOH is in excess nor H

_{2}SO

_{4}therefore solution will be neutral and hence

**pH= 7**

**
**