Which of the following would decrease the pH of 25 cm`""^(2)` of a 0.01 M solution of HCl?

Options

A. The addition of `25 cm^(3) 0.005` M hydrochloric acid
B. The addition of `25 cm^(3) " of " 0.02 ` M hydrochloric acid
C. The addition of magnesium metal
D. None of these

Answer

B

Explanation

Given concentration of [HCl] = 0.01 M
`pH = - log 10^(-2) = 2 `
In option (b)
Concentration of [HCl] = 0.02 M
M.eg = `0.02 xx 25 = 0.5` M.eg of HCl
M.eg in given condition = `0.01 xx 25 = 0.25 `
Total M.eg in solution = `0.25 + 0.50 = 0.75`
`= (0.75)/(25 +25) = 1.5 xx 10^(-2)N `
Then new pH will be
`= - log 1.5 xx 10^(-2) = 2 - log 1.5`
Which is less than 2
So we can vary that by adding `25cm^(3) " of " 0.2 M ` HCl
then pH of `25 cm^(3)` of `0.01` M decrease .