What is the pOH of a solution prepared by dissolving 0.814 g of KOH(s) in 7.70 L of water?
A.7.000
B.1.838
C.12.162
D.2.725
E.11.275
Molar mass of KOH = 1*MM(K) + 1*MM(O) + 1*MM(H)
= 1*39.1 + 1*16.0 + 1*1.008
= 56.108 g/mol
mass of KOH = 0.814 g
we have below equation to be used:
number of mol of KOH,
n = mass of KOH/molar mass of KOH
=(0.814 g)/(56.108 g/mol)
= 1.451*10^-2 mol
volume , V = 7.70 L
we have below equation to be used:
Molarity,
M = number of mol / volume in L
= 1.451*10^-2/7.7
= 1.884*10^-3 M
SO,
[OH-] = 1.884*10^-3 M
we have below equation to be used:
pOH = -log [OH-]
= -log (1.884*10^-3)
= 2.725
Answer: 2.725
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