Question

5.6 g of KOH are dissolved in 100 litre aqueous solution . Calculate the ph at 353k . if Kw = 1x10 ^-12​

Answer 1

hope this example can help u :-

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 10-2As molar mass of KOH = 56 uConcentration of solution = (moles of KOH / vol. of solution) x 1000= (1.0 x 10-2 / 200) x 1000= 5.0 x 10-2 MAs KOH is completely ionized therefore,[OH–] = [KOH] = 5.0 x 10-2 M[H3O+] = Kw / [OH–]= 1 x 10-14 / 5.0 x 10-2= 2.0 x 10-13pH = – log [H3O+] = – log (2.0 x 10-13)= – log 2.0 + 13 = 0.301 + 3 = 12.699Hence, the pH of the 0.561 g of KOH dissolved in 200 mL of solution is equal to 12.699