Please explain in detail-
Please explain in detail- 66. 100 mL of 0.1. M NaOH solution is titrated with 100 mL of 0.05 M H2S04 solution. The pH of the resulting solution is : (For H2SOg,Kal = oo,Kä2 = 10-2) (b) 7.2 (c) 7.4 (d) 6.8

Dear Student,

Normality of NaOH, N_{1} = Molarity of NaOH \times 1 = 0.1 N Normality of H_{2} {SO}_{4}, N_{2} = Molarity of H_{2} {SO}_{4} \times 2 = 0.05 \times 2 = 0.01 N Now, Equivalent of NaOH = N_{1} V_{1} = 0.1 L \times 0.1 N = 0.01 Equivalent of H_{2} {SO}_{4} = N_{2} V_{2} = 0.01 L \times 0.1 N = 0.01 Since Equivalent of NaOH = Equivalent of H_{2} {SO}_{4} So, complete neutralisation takes place and hence, pH = 7 The correct answer is (a)

Please explain in detail-Please explain in detail- 66. 100 mL of 0.1. M NaOH solution is titrated with 100 mL of 0.05 M H2S04 solution. The pH of the resulting solution is : (For H2SOg,Kal = oo,Kä2 = 10-2) (b) 7.2 (c) 7.4 (d) 6.8