Question QUESTION 28 A 20.00-ml sample of 0.3000 MHBr is titrated with 0.15 M NaOH. What is the pH of the solution after 30.00 ml of NaOH have been added to the acid? Horta-NaOH(aq) NaBrag). H2Otag)

We first calculated moles of acid and bases ,then calculated excess reagent using that calculated it's molar concentration . Then using that calculated pH.   \mathrm{HBr}+\mathrm{MaOH} \longrightarrow \mathrm{MaBr}+\mathrm{H}_{2} \mathrm{O}\longrightarrow Males of Aid H B r= Molanity x Volumes in liters\begin{array}{l}=0.3 \frac{\mathrm{mol}}{\%} \times 20 \times 10^{-3} \\=6 \times 10^{-3} \text { moles. }\end{array}\longrightarrow Moles\text { of } \begin{aligned}\mathrm{MNOH} & =0.15 \text { mul } \times 30 \times 10^{-3} \mathrm{Y} \\& =4.5 \times 10^{-3} \text { moles. }\end{aligned}So pearly me have emess of \mathrm{H}^{+}in solusian.Swhtion retal whame =(30+20) \mathrm{ml}=50 m t=50 \times 10^{-3} \mathrm{~L} \text {. }Evens \mathrm{H}^{+}in solution = Moles of Aird-moles of Base\begin{array}{l}=10^{-3}(6-4.5) \\=10^{-3} \times 1.5 \text { moles }\end{array}\text { So cementration of } \begin{aligned}H^{+} & =\frac{1.5 \times 10^{-3}}{50 \times 10^{-5}} \mathrm{moles} \\& =0.03 \mathrm{M}\end{aligned}So\begin{aligned}P H & =-\log [H+] \\& =-\log [0.03] \\& =1.523\end{aligned} ...