Question

What mass of KX (M_m= 150 g/mol) is required to prepare 331 mL of a pH = 12.68 solution? The pKa of HX is 11.75

g KX

Answer 1

mass of KX = 22.6 g

Explanation

pK_a of HX = 11.75

pK_b of X^- = 14 - pK_a of HX

pK_b of X^- = 14 - 11.75

pK_b of X^- = 2.25

K_b = 10^-pK_b

K_b = 10^-2.25

K_b = 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 / K_b) + [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