Question

1ml of 0.1NHCl is added to 999ml solution of NaCl. The pH of the resulting solution will be:

Answer 1

V_HCl ​ =0.001L,  N_HCl ​ =0.1N 

V_NaCl ​ =0.999L 

V_solution ​ =V_HCl ​ +V_NaCl ​ =1 

∴ Normality of HCl in a resulting solution=0.1×0.001/1 ​ =0.0001N=[H⁺] 

Addition of NaCl does not effect the pH of solution as it is a salt & is neutral. 

So, pH=−log[H⁺] = −log[0.0001] = 4