Question

What is the pOH of a solution prepared by dissolving 0.835 g of KOH(s) in 7.70 L of water?

A.2.714

B.1.827

C.7.000

D.11.286

E.12.173

Answer 1

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 = 0.835 g

we have below equation to be used:

number of mol of KOH,

n = mass of KOH/molar mass of KOH

=(0.835 g)/(56.108 g/mol)

= 1.488*10^-2 mol

volume , V = 7.70 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 1.488*10^-2/7.7

= 1.933*10^-3 M

[OH-] = [KOH] = 1.933*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.933*10^-3)

= 2.714

Answer: A