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 0.02 M HCl
C. Addition of 150 mL of 0.02 M HCl
D. Addition of 5 mL of 1 M HCl

Answer

D

Explanation

For 0.01 M HCl, `|H^(+)| =10^(-2), pH=2`
`(A) N_(1)V_(1)+N_(2)V_(2)=N_(3)(V_(1)+V_(2))`
`50 xx 0.01+50 xx 0.01 =N_(3)(50+50)`
or `N_(3)=(.5 +.5)/(100)=10^(-2),pH=2` (No change)
`(B) 50 xx 0.01 +c0 xx 0.002=N_(3)(50+50)`
or `N_(3) =(0.5 +0.1 )/(100)=6 xx 10^(-3)`
`pH= - log (6 xx 10^(-3))=2.22 (`pH increases)
`(C ) 50 xx 0.01 +150 xx 0.002 = N_(3)(50+150)`
or `N_(3)=(.5 +.3)/(200)=4 xx 10^(-3)`
`pH =-log (4 xx 10^(-3))=2.39 pH (pH ` increases )
`(D) 50 xx 0.01 +5 xx 1 =N_(3)(50+5)`
or `N_(3)=(0.5 +5)/(55)=(5.5)/(5)=10^(-1)`
`pH =1 (`pH decreases )