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)= 812 mg
= 0.812 g
use:
number of mol of Ba(OH)2,
n = mass of Ba(OH)2/molar mass of Ba(OH)2
=(0.812 g)/(1.713*10^2 g/mol)
= 4.74*10^-3 mol
volume , V = 1.3*10^2 mL
= 0.13 L
use:
Molarity,
M = number of mol / volume in L
= 4.74*10^-3/0.13
= 3.646*10^-2 M
So,
[Ba(OH)2] = 3.646*10^-2 M
So,
[OH-] = 2*[Ba(OH)2]
= 2*3.646*10^-2 M
= 7.292*10^-2 M
use:
pOH = -log [OH-]
= -log (7.292*10^-2)
= 1.1372
use:
PH = 14 - pOH
= 14 - 1.1372
= 12.8628
Answer: 12.863
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