600 , mL of 0.01 , M , HCl is mixed with 400 | vedclass

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(B) Total millimoles of $H^{+} = (600 	imes 0.01) + (400 	imes 0.01 	imes 2) = 6 + 8 = 14 \, mmol$. Total volume $= 600 + 400 = 1000 \, mL$. Concen

Vedclass · Accepted Answer

$186$

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(B) Total millimoles of $H^{+} = (600 	imes 0.01) + (400 	imes 0.01 	imes 2) = 6 + 8 = 14 \, mmol$. Total volume $= 600 + 400 = 1000 \, mL$. Concentration of $[H^{+}] = \frac{14 \, mmol}{1000 \, mL} = 0.014 \, M = 1.4 	imes 10^{-2} \, M$. $pH = -\log[H^{+}] = -\log(1.4 	imes 10^{-2}) = 2 - \log(1.4) = 2 - \log(14/10) = 2 - (\log 14 - \log 10) = 2 - (\log 2 + \log 7 - 1) = 2 - (0.30 + 0.84 - 1) = 2 - 0.14 = 1.86$. Thus,$pH = 1.86 = 186 	imes 10^{-2}$.

$600 \, mL$ of $0.01 \, M \, HCl$ is mixed with $400 \, mL$ of $0.01 \, M \, H_2SO_4$. The $pH$ of the mixture is $....... \times 10^{-2}$. (Nearest integer) [Given $\log 2 = 0.30, \quad \log 3 = 0.48, \quad \log 5 = 0.69, \quad \log 7 = 0.84, \quad \log 11 = 1.04$]

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