MCQ
Given $pH $ of a solution $ A$ is $3$ and it is mixed with another solution $B$ having $pH\, 2$. If both mixed then resultant $pH$ of the solution will be
- A$3.2$
- ✓$1.9$
- C$3.4$
- D$3.5$
$[H+]_A = 10^{-3} M.$
$pH $ of the solution $B = 2$
$[H^+]_B =10^{-2} M$
$[H^+] = 10^{-3} + 10^{-2} = 10^{-3} + 10 × 10^{-3} = 11 \times 10^{-3}$.
$pH = -log(11 \times 10^{-3}) = 3 -log 11$ $= 3 -1.04 = 1.95$
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