What will be the pH of a solution formed by m | vedclass

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(A) Millimoles of $HCl = 0.10 \, M 	imes 40 \, mL = 4 \, mmol$.Millimoles of $NaOH = 0.45 \, M 	imes 10 \, mL = 4.5 \, mmol$.Since $NaOH$ is in exce

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$12$

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(A) Millimoles of $HCl = 0.10 \, M 	imes 40 \, mL = 4 \, mmol$.Millimoles of $NaOH = 0.45 \, M 	imes 10 \, mL = 4.5 \, mmol$.Since $NaOH$ is in excess,the remaining $OH^-$ ions after neutralization are $4.5 - 4 = 0.5 \, mmol$.Total volume of the solution $= 40 \, mL + 10 \, mL = 50 \, mL$.Concentration of $[OH^-] = \frac{0.5 \, mmol}{50 \, mL} = 0.01 \, M = 10^{-2} \, M$.$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.$pH = 14 - pOH = 14 - 2 = 12$.

What will be the $pH$ of a solution formed by mixing $40 \, mL$ of $0.10 \, M \, HCl$ with $10 \, mL$ of $0.45 \, M \, NaOH$?

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