Equal volumes of three acid solutions of pH 3 | vedclass

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(D) The $pH$ is defined as $pH = -\log[H^{+}]$,which implies $[H^{+}] = 10^{-pH}$.For the three solutions:$[H^{+}]_1 = 10^{-3} \ M$$[H^{+}]_2 = 10^{-4

Vedclass · Accepted Answer

$3.7$

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(D) The $pH$ is defined as $pH = -\log[H^{+}]$,which implies $[H^{+}] = 10^{-pH}$.For the three solutions:$[H^{+}]_1 = 10^{-3} \ M$$[H^{+}]_2 = 10^{-4} \ M$$[H^{+}]_3 = 10^{-5} \ M$Assuming equal volumes $V$ for each solution,the total volume of the mixture is $3V$.The total moles of $H^{+}$ ions are $n_{total} = (10^{-3} 	imes V) + (10^{-4} 	imes V) + (10^{-5} 	imes V) = V(10^{-3} + 0.1 	imes 10^{-3} + 0.01 	imes 10^{-3}) = V(1.11 	imes 10^{-3})$.The final concentration is $[H^{+}]_{mix} = \frac{n_{total}}{3V} = \frac{1.11 	imes 10^{-3}}{3} = 0.37 	imes 10^{-3} \ M$.Converting this to the form $	imes 10^{-4} \ M$,we get $3.7 	imes 10^{-4} \ M$.

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 in terms of $\times 10^{-4} \ M$?

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