Question

Please answer these question:

--- Calculate the Ka2 for oxalic acid (found in rhubarb and spinach) (H2C204) if a 0.35 M solution of Na2C204 has a pH of 8.87. -- Calculate the molarity of a solution of nicotine, a weak base, if the solution has a pH = 10.72. Kb = 7.4 x 10-7

Answer 1

part 1

Na₂C₂O₄ is salt of strong base ( NaOH) and weak acid (NaHC₂O₄) .

given concentration of salt (C_salt) = 0.35 M

hence,

pH = 7 + $\frac{1}{2}$ ( pKa₂ + logC_salt )

pH = 7 + $\frac{1}{2}$ ( pKa₂ + log 0.35)

or, 8.87 = 7 + $\frac{1}{2}$ ( pKa₂ - 0.456)

or, pKa₂ - 0.456 = 2 × (8.87 - 7)

or, pKa₂ - 0.456 = 3.74

or, pKa₂ = 3.74 + 0.456 = 4.196.

hence Ka₂ of oxalic acid  = 10^-4.196 = 6.36×10^-5

so,

part 2

given , K_b = 7.4×10^-7 , hence,

pKb = - logKb = - log (7.4×10^-7) = 6.13

pH = 10.72

for weak base

pH =14 - $\frac{1}{2}$ ( pKb - logC)

or, 10.72 = 14 - $\frac{1}{2}$ ( 6.13 - log C)

or, -$\frac{1}{2}$ (6.13 - logC) = (10.72 - 14)

or, - (6.13 - log C) = - (3.28×2)

or, log C - 6.13 = - 6.56

or, log C = (6.13 - 6.56)

or, log C = - 0.43

or, C = 10^(-0.43)

or, C = 0.37 M.

concentration of nicotine = 0.37 M.