100 ml of 0.1 N NaOH is mixed with 100 ml of 0.1 N H₂SO₄.The pH of the resultant solution is??

NaOH  volume = 100ml or 0.1L 
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]

H₂SO₄ volume = 100ml or 0.1L 
Normality= 0.1 N
number of electron change (n) for H₂SO₄  = 2
Thus molarity = 0.05 M  ( since N = M $\times$n)
Moles of H₂SO₄ = 0.1$\times$0.1 = 0.005 mol

0.01 mole of NaOH exactly neutralizes 0.005 mole of H₂SO₄ according to the reaction
NaOH + H₂SO₄ $\to$Na₂SO₄ + H₂O
Thus neither NaOH is in excess nor H₂SO₄ therefore solution will be neutral and hence
pH= 7