The pH of a solution obtained by mixing 50 mL of 1N HCl and 30 mL of 1N NaOH is `[log 2.5 =0.3979]`

Options

A. 3.979
B. 0.6021
C. 12.042
D. 1.2042

Answer

B

Explanation

No. of milli equivalent of `NaOH = 30 xx 1 = 30`
No. of milli equivalent of `HCl = 50 xx 1 = 50`
`:.` No. of milli equivalent of HCl left after titration = 50 - 30 = 20.
Total volume of the mixture = 50 + 30 = 80 ml
i.e. 20 milli equivalent or 0.02 equivalent of HCl are present in 80 ml.
`:.` 250 milli equivalent or 0.25 equivalent of HCl are present in 1000 ml or 1 litre.
i.e., `0.25 N HCl ~~ 0.25 N NaOH` (Monobasic)
So, `[H^(+)] = -log_(10)[2.5 xx 10^(-1)] rArr pH = 1 - log_(10)[0.25]`
`pH = -log_(10)[2.5 xx 10^(-2) rArr pH = 1 - log_(10)2.5`
`pH = 1 - 0.3979 rArr pH = 0.6021`.