Consider 1 L of solution.
Molar mass of Ba(OH)2,
MM = 1*MM(Ba) + 2*MM(O) + 2*MM(H)
= 1*137.3 + 2*16.0 + 2*1.008
= 171.316 g/mol
mass(Ba(OH)2)= 3.77 mg
= 0.00377 g
use:
number of mol of Ba(OH)2,
n = mass of Ba(OH)2/molar mass of Ba(OH)2
=(3.77*10^-3 g)/(1.713*10^2 g/mol)
= 2.201*10^-5 mol
volume , V = 1 L
use:
Molarity,
M = number of mol / volume in L
= 2.201*10^-5/1
= 2.201*10^-5 M
So,
[OH-] = 2[Ba(OH)2]
= 2*2.201*10^-5 M
= 4.402*10^-5 M
use:
pOH = -log [OH-]
= -log (4.402*10^-5)
= 4.3563
use:
PH = 14 - pOH
= 14 - 4.3563
= 9.6437
Answer: 9.64
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