Question

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

Answer 1

If 0.561 g of KOH is dismissed in water to give 200 mL of water at 298 K.

Figure the absorptions of potassium, hydrogen and hydroxyl ions.

Then the solution will be as follows [KOH(aq)] = 0.561 / (1/5)g/L

= 2.805 g/L

= 2.805 x 1/56.11 M

= 0.05M

 

KOH(aq)  → K+(aq)  +  OH-(aq)

[OH-]  = 0.05M =  [K+]

[H-] [H+]  =  Kw

[H+] Kw / [OH-]

= 10-14 / 0.05 = 2x10-13 M

∴ pH = 12.70

Answer 2

Answer:

If 0.561 g of KOH is dismissed in water to give 200 mL of water at 298 K.

Figure the absorptions of potassium, hydrogen and hydroxyl ions.

Then the solution will be as follows [KOH(aq)] = 0.561 / (1/5)g/L

= 2.805 g/L

= 2.805 x 1/56.11 M

= 0.05M

 

KOH(aq)  → K+(aq)  +  OH-(aq)

[OH-]  = 0.05M =  [K+]

[H-] [H+]  =  Kw

[H+] Kw / [OH-]

= 10-14 / 0.05 = 2x10-13 M

∴ pH = 12.70