For solution "A" ,"[H^+](M_1)=10^{-5} mol\/L"
Volume"(V_1) =14\\ cm^3=14 \\times 10^{-3} dm^3=14 ml"
For solution B,"[H^+]=10^{-3} mol\/L"
Volume"(V_2)=14 cm^3=14 \\ ml"
After mixing,Volume will be"(V_3)=14+14=28 ml"
Using molarity equation of mixing,
"M_1V_1+M_2V_2=M_3V_3"
"10^{-5}\\times 14+10^{-3}\\times 14=M_3\\times 28"
"M_3=\\frac{10^{-3}+10^{-5}}{2}=\\frac{101}{200000}" mol/L
pH of solution"=-log(M_3)=3.30"
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