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 ?
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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
Answered by
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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
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