Which of the following will decrease the pH of a 50 ml solution of `0.01 M HCI` ?

Options

A. Addition of 50 ml of 0.01 M HCl
B. Addition of 50 ml of 0.002 M HCl
C. Addition of 150 mL of 0.002 M HCl
D. Addition of metallic zinc

Answer

D

Explanation

For 0.01 M HCl , pH=-log `10^(-2)` =2
For (a) : `N_1V_1+N_2V_2=N_3(V_1+V_2)`
`50xx0.01+50xx0.01=N_3(100)`
`N_3=(0.5+0.5)/100=10^(-2)`
pH=-log `(10^(-2))` =2 (No change in pH)
For (b): 50 x 0.01 + 50 x 0.002 = `N_3xx100`
or `N_3=(0.5+0.1)/100=6xx10^(-3)`
`pH=-log (6xx10^(-3))` =2.22 (pH increased )