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Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the $H^+$ ion concentration in the mixture ?
A
$3.7 \times 10^{-3}$ M
B
$1.11 \times 10^{-3}$ M
C
$1.11 \times 10^{-4}$ M
D
$3.7 \times 10^{-4}$ M
Detailed Solution
$pH = -\log[H^+]$, so $[H^+] = 10^{-pH}$
$[H^+]$ of solution 1 $= 10^{-3}$ M
$[H^+]$ of solution 2 $= 10^{-4}$ M
$[H^+]$ of solution 3 $= 10^{-5}$ M
Let the volume taken in each case be 1 L. Total moles of $H^+$ $= 10^{-3} + 10^{-4} + 10^{-5}$
$= 10^{-3}(1 + 1 \times 10^{-1} + 1 \times 10^{-2})$
$= 10^{-3}\left(\dfrac{100 + 10 + 1}{100}\right) = 10^{-3}\left(\dfrac{111}{100}\right) = 1.11 \times 10^{-3}$ mol
Total volume of the mixture = 3 L
Therefore, $H^+$ ion concentration in the mixture $= \dfrac{1.11 \times 10^{-3}}{3} = 3.7 \times 10^{-4}$ M
$[H^+]$ of solution 1 $= 10^{-3}$ M
$[H^+]$ of solution 2 $= 10^{-4}$ M
$[H^+]$ of solution 3 $= 10^{-5}$ M
Let the volume taken in each case be 1 L. Total moles of $H^+$ $= 10^{-3} + 10^{-4} + 10^{-5}$
$= 10^{-3}(1 + 1 \times 10^{-1} + 1 \times 10^{-2})$
$= 10^{-3}\left(\dfrac{100 + 10 + 1}{100}\right) = 10^{-3}\left(\dfrac{111}{100}\right) = 1.11 \times 10^{-3}$ mol
Total volume of the mixture = 3 L
Therefore, $H^+$ ion concentration in the mixture $= \dfrac{1.11 \times 10^{-3}}{3} = 3.7 \times 10^{-4}$ M