Odpowiedź:
[tex]n_{KOH}=C_m*V=0,1*0,15=0,015mol\\pH=3 \implies [H+]=10^{-3}=0,001mol/dm^3\\V_r=0,15dm^3+V_{HCl}\\V_{HCl}=\frac{n_{HCl}}{0,2mol/dm^3}\\V_r=0,15dm^3 + \frac{n_{HCl}}{0,2mol/dm^3}\\{[H^+]}=\frac{n_{HCl}-0,015}{0,15dm^3 + \frac{n_{HCl}}{0,2mol/dm^3}}=0,001mol/dm^3\\0,001({0,15 + \frac{n_{HCl}}{0,2}})=n_{HCl}-0,015/*1000\\{0,15 + \frac{n_{HCl}}{0,2}}=1000n_{HCl}-15/*0,2\\0,03+n_{HCl}=200n_{HCl}-3\\3,03=199n_{HCl}\\n_{HCl}\approx0,0152mol\\V_{HCl}=\frac{0,0152}{0,2}=0,076dm^3[/tex]
Wyjaśnienie: