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Answer= pH of the solution is 10.87 Ansi molarity of \mathrm{HBr}=0.3 \mathrm{~m}volume of \mathrm{ABr}=20 \mathrm{ml}Malavity =\frac{\text { no. of moles } \times 1000}{\text { volume in mL }}no. of molen of \mathrm{HBr}=\frac{\text { molarity } x \text { volume }}{1000}\begin{array}{l}=\frac{0.3 \mathrm{~m} \times 20 \mathrm{ml}}{1000} \\\Rightarrow 6.0 \times 10^{-3} \text { molen } .\end{array}\rightarrow Molarity of Naon =0.15 \mathrm{~m}volume of vaon =40.3 \mathrm{~mL}no. of molen of vaon = malarity x volume1000\begin{array}{l}=\frac{0.15 \mathrm{~m} \times 40.3 \mathrm{~mL}}{1000} \\\Rightarrow 6.045 \times 10^{-3} \text { moles }\end{array}According to reacion \mathrm{HCl}+\mathrm{NaOn} \longrightarrow \mathrm{NaCl}+\mathrm{n}_{2} \mathrm{O} i mole HCl leats writh I mole NaOn.\rightarrow 6.0 \times 10^{-3} mole HU recuts with 6.0 \times 10^{-3} mole of Naon\rightarrow No. of moler of Naon left unrecuted\begin{array}{l}=6.045 \times 10^{-3} \text { mole }-6.0 \times 10^{-3} \text { mole } \\=0.045 \times 10^{-3} \text { mole } \\=4.5 \times 10^{-5} \text { moles. }\end{array}volume of solution = valume of x t t volume of Naon\begin{array}{l}=20 m l+40.3 m L \\=60.3 m L \\=60.3 \times 10^{-3} \text { litre }\end{array}molarity of unbuated vaon =\begin{array}{l}=\frac{\text { no. of molen of vaon unreated }}{\text { volume of solution }} \\=\frac{4.5 \times 10^{-5} \text { molen }}{60.3 \times 10^{-3} \text { litre }} \\\Rightarrow 7.46 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\end{array}As Naon is strong electo lyte so it dissocicuten completely amd concentration of on-ion is 7.46 \times 10^{-4} \mathrm{~mol} / \mathrm{h}Now\begin{array}{l}\text { pon }=-\log [0 n] \\\text { pon }=-\log \left(7.46 \times 10^{-4}\right) \\\text { pon }=-\log (7.46)+\left(-\log 10^{-4}\right) \\\text { pon }=-0.87+(-1 \times-4 \log 10) \\\text { pon }=-0.07+4 \\\text { pon }=3.13\end{array}As p n+p o n=14\begin{array}{l}p n=14-p 0 n \\p n=14- \\p n=10.87 .\end{array}A m \rightarrow p n of solution is 10.87 . ...