On adding which of the following the pH of 20mL of 0.1N HCl will not alter

Options

A. 1 mL of 1 HCl
B. 20 mL of distilled water
C. 1 mL of 0.1 N NaOH
D. 500 mL of HCl of pH =1

Answer

D

Explanation

`0.1 N " HCl " -= 0.1 M "HCl " rArr pH = 1 `
No. moles in 20 mL of HCl
` = 0.1 xx 20 xx10^(-3) = 2 xx 10^(-3)`
(d) `[H^(+)] = 10^(-pH) =10^(-1) = 0.1 M `
No. of moles in 500 mL of HCl
`= 0.1 xx 500 xx 10^(-3) = 50 xx 10^(-3)`
Total no . of moles in solution = ` 2xx 10^(-3) + 50 xx 10^(-3)`
Total volume of solution = `20 + 500 = 520 mL `
New `[H^(+)] = (10^(-3)(2+50))/(520 xx 10^(-3)) = 0.1 `
`pH = - log [H^(+)] = - log (0.1) = 1 `