What mass of KX (Mm= 150 g/mol) is required to prepare 331 mL of a pH = 12.68 solution? The pKa of HX is 11.75
g KX
mass of KX = 22.6 g
Explanation
pKa of HX = 11.75
pKb of X- = 14 - pKa of HX
pKb of X- = 14 - 11.75
pKb of X- = 2.25
Kb = 10-pKb
Kb = 10-2.25
Kb = 5.62 x 10-3
pH = 12.68
pOH = 14 - pH
pOH = 14 - 12.68
pOH = 1.32
[OH-] = 10-pOH
[OH-] = 10-1.32
[OH-] = 0.04786 M
initial concentration of KX = ([OH-]2 / Kb) + [OH-]
initial concentration of KX = [(0.04786 M)2 / 5.62 x 10-3] + (0.04786 M)
initial concentration of KX = 0.4555 M
volume of solution = 331 mL = 0.331 L
moles of KX = (initial concentration of KX) * (volume of solution in Liter)
moles of KX = (0.4555 M) * (0.331 L)
moles of KX = 0.151 mol
mass of KX = (moles of KX) * (molar mass KX)
mass of KX = (0.151 mol) * (150 g/mol)
mass of KX = 22.6 g
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