Question

Consider the titration of $50mL$ of $0.1M$ $HBr$ with $0.1M$ $KOH$. Calculate $pH$ after $49mL$ of the base has been added to the $50mL$ of $HBr$.

Answer 1

$50$ ml of $0.1M$ $HBr=\frac{50\times 0.1}{1000}=5\times {10}^{-3}$ moles

$49$ ml of $0.1M$ $KOH=\frac{49\times 0.1}{1000}=4.9\times {10}^{-3}$ moles

$\therefore$ Excess moles of $HBr$ will be $0.1\times {10}^{-3}$ moles.

$\therefore$ $pH=-log(\frac{0.1\times {10}^{-3}}{99}\times 1000)\approx 3$.