Question

What would be the PH of a solution obtained by mixing 100 mL of 0.1 (N) HCL and 9.9 mL of 1.0 (N) NAOH ?​

Answer 1

Answer

Step by step solution

Solution

100 ml of 0.1 N HCl = 100×0.1 milli eq. = 10 milli eq.

9.9 ml of 1 N NaOH =9.9×1 milli eq. = 9.9 milli eq.

∴ HCl left unneutralized =10−9.9=0.1 milli eq.

Volume of solution = 100 + 9.9 = 109.9 ml

∴ Normality pf HCl in resulting solution =0.1109.9≈0.1110=0.09×10−4N=9.09×10−4M

As HCl completely ionizes as HCl+H2O→H3O++Cl−

∴[H3O+]=[HCl]=9.09×10−4M

pH=−log[H3O+]=−log(9.09×10−4)=−[log9.09+log10−4]=−[0.9546−4]

=3.0454=3.05