1.

The pH of a solution obtained by mixing equal volumes of (N)/(10)NaOH and (N)/(20) HCl

Answer»

13.4
12.4
7.6
1.6

Solution :`N_(1)V_("base") - N_(2)V_("2 acid") = N_(R) (V_(1)+V_(2))`
`(1)/(10) xx V -(1)/(20) xx V = N_(R)xx 2 xx V`
`(0.05)/(2) = N_(R) rArr N_(R) = 0.025`.
`[H^(+)] = 0.025 M`
`PH = -log[0.025] = 1.6`.


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