Question

If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

Answer 1

[KOHaq]
= $\frac{0.561}{\frac{1}{5}}$ g / L
= 2.805 g / L
= 2.805 × $\frac{1}{56.11}$ M
= .05M
KOH(aq) $\rightarrow$ K+(aq) + OH-(aq)
= [OH-] = .05M
= [K+] = [H+][H-]
= Kw = $\frac{[H^+] K_w}{[OH^-] }$
= $\frac{10^{-14}}{0.05}$= 2 × 10-13 M
∴ pH = 12.70