At , pure water has . If of is added to of at , what will be the exact pH of the resulting solution? — Ionic Equilibrium Chemistry Question
Question
At $90^\circ C$, pure water has $[H^+] = 10^{-6} \text{ M}$. If $100 \text{ mL}$ of $0.2 \text{ M } HNO_3$ is added to $20 \text{ mL}$ of $1.0 \text{ M } NaOH$ at $90^\circ C$, what will be the exact pH of the resulting solution?
💡 Solution & Explanation
Moles of $H^+$ from $HNO_3 = 100 \text{ mL} \times 0.2 \text{ M} = 20 \text{ mmol}$. Moles of $OH^-$ from $NaOH = 20 \text{ mL} \times 1.0 \text{ M} = 20 \text{ mmol}$. Since the millimoles of strong acid and strong base are perfectly equal, complete neutralization occurs, resulting in a perfectly neutral solution. At $90^\circ C$, the neutral pH is dictated by the water auto-ionization, which is $-\log(10^{-6}) = 6$.