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 5 mL of 1 M HCl

Answer

D

Explanation

For 0.01 M HCl, `[H^(+)]=10^(-2)M`, pH=2
(a) `N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))`
`50xx0.01+50xx0.01=N_(3)(50+50)`
or `N_(3)=(0.5+0.5)/(100)=10^(-2), pH=2` (No change)
(b) `50xx0.01+50xx0.02=N_(3)(50+50)`
or, `N_(3)=(0.5+0.1)/(100)=6xx10^(-3)`
`pH=-log(6xx10^(-3))=2.22` (pH increases)
(c) `50xx0.01+150xx0.002=N_(3)(50+150)`
or, `N_(3)=(0.5+0.3)/(200)=4xx10^(-3)`
`pH=-log(4xx10^(-3))=2.39` (pH increases)
(d) `50xx0.01+5xx1=N_(3)(50+5)`
or, `N_(3)=(0.5+5)/(55)=(55)/(5)=10^(-1)`
pH = 1 (pH decreases)