Question

30 cc of M/3 HCl, 20cc of
M/2 HNO3 and 40cc of
M/4 NaOH solutions are
mixed and the volume
was made up to 1dm3.
The pH of the resulting
solution is​

Answer 1

Answer:

2

Explanation:

Total milli equivalent of h+ = 20 x 1/2 + 30 x 1/3 = 20

Total milliequivalent of O$H^{-}$ = 40 x 1/4 = 10

Thus, milliequivalents of H+ left = 20 - 10 =10

Total volume of solution = 1 dm³ = 1000ml

===> [H+] = $\frac{10}{1000}$ = $10^{-2}$

===> pH = -log[H+] = -log [$10^{-2}$] = 2

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