Question

10ml of a solution having pH=4 is added with 990mL of 0.1M NaCl solution. pH of resulting solution is

Answer 1

Volume of HCl = 10 mL or 0.01 L

Normality of HCl = 0.1 N

Number of gram-equivalents of HCl = Normality × Volume

= 0.1 × 0.01

= 0.001

Volume of solution = Volume of NaCl + Volume of HCl

= 990 + 10

= 1000 mL

= 1 L

As NaCl is salt it will not contribute towards the pH of the solution.

Normality of HCl in resulting solution = 0.001 / 1

= 0.001 N

pH = -log[H+]

= - log(0.001)

= 3

if there is something wrong then kindly rectify it

if there is something wrong then kindly rectify it