Calculate the exact pH of a solution formed by mixing of , of , of , and of pure . (Given ). Find th — Ionic Equilibrium Chemistry Question
Question
Calculate the exact pH of a solution formed by mixing $50 \text{ mL}$ of $0.1 \text{ M } HCl$, $25 \text{ mL}$ of $0.1 \text{ M } H_2SO_4$, $25 \text{ mL}$ of $0.2 \text{ M } HNO_3$, and $100 \text{ mL}$ of pure $H_2O$. (Given $\log 7.5 = 0.875$). Find the exact value of the resulting pH.
💡 Solution & Explanation
All listed compounds are strong acids. Calculate total millimoles of $H^+$: from $HCl = 50 \times 0.1 = 5 \text{ mmol}$; from $H_2SO_4 = 25 \times 0.1 \times 2 = 5 \text{ mmol}$; from $HNO_3 = 25 \times 0.2 = 5 \text{ mmol}$. Total $H^+$ is $15 \text{ mmol}$. Total volume is $50 + 25 + 25 + 100 = 200 \text{ mL}$. The final molarity is $[H^+] = 15/200 = 0.075 \text{ M}$. The pH is $-\log(0.075) = -\log(7.5 \times 10^{-2}) = 2 - \log 7.5 = 2 - 0.875 = 1.125$.