Calculate the pH of the 0.561 g of KOH dissolved in 200 mL of solution.

0.561 g of KOH dissolved in 200 mL of solution will have:
Moles of KOH = 0.561 / 56 = 1.0 x 10^-2
As molar mass of KOH = 56 u
Concentration of solution = (moles of KOH / vol. of solution) x 1000
= (1.0 x 10^-2 / 200) x 1000
= 5.0 x 10^-2 M
As KOH is completely ionized therefore,
[OH^–] = [KOH] = 5.0 x 10^-2 M
[H₃O⁺] = K_w / [OH^–]
= 1 x 10^-14 / 5.0 x 10^-2
= 2.0 x 10^-13
pH = – log [H₃O⁺] = – log (2.0 x 10^-13)
= – log 2.0 + 13 = 0.301 + 3 = 12.699

Hence, the pH of the 0.561 g of KOH dissolved in 200 mL of solution is equal to 12.699