The pH of the solution obtained by mixing 100 | vedclass

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(B) For solution $I, pH=3$ $\Rightarrow [H^{+}] = 10^{-3} \ M$ $V_{1} = 100 \ mL$ For solution $II, pH = 4$ $\Rightarrow [H^{+}] = 10^{-4} \ M$ $V_{2}

Vedclass · Accepted Answer

$4-\log 2.8$

Vedclass · Answer

(B) For solution $I, pH=3$ $\Rightarrow [H^{+}] = 10^{-3} \ M$ $V_{1} = 100 \ mL$ For solution $II, pH = 4$ $\Rightarrow [H^{+}] = 10^{-4} \ M$ $V_{2} = 400 \ mL$ Concentration of resulting solution $M = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}$ $M = \frac{10^{-3} 	imes 100 + 10^{-4} 	imes 400}{100 + 400}$ $M = \frac{0.1 + 0.04}{500} = \frac{0.14}{500} = 2.8 	imes 10^{-4} \ M$ $[H^{+}] = 2.8 	imes 10^{-4} \ M$ $pH = -\log [H^{+}] = -\log (2.8 	imes 10^{-4})$ $pH = 4 - \log 2.8$

The $pH$ of the solution obtained by mixing $100 \ mL$ of a solution of $pH=3$ with $400 \ mL$ of a solution of $pH=4$ is

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