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The pH of a solution obtained by mixing ...

The pH of a solution obtained by mixing 100mL of 0.4HCl and 100 mL of 0.2 N NaOH is :

A

`-log 2`

B

`-log 0.2`

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
C
`N_(1)V_("1 acid") - N_(2) V_("2 base") = N_(R)(V_(1) + V_(2))`
`0.4 xx 50 0.2 xx 50 = N_(R) xx 100, N_(R) = 0.1 therefore [H^(+)] = 0.1 M`
`pH = -log[H^(+)] = -log 0.1 = 1`
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