In: Chemistry
How many grams of sodium lactate (MW = 112 g/mol) are required to prepare 200 ml solution at 1.0 M? Given pKa of lactic acid = 3.86, what is the pH of this solution?
number of moles of sodium Lactate = M*V
= 1M * 0.2 L
= 0.2 mol
mass of sodium lactate = MW * number of moles
= 112 g/mol * 0.2 mol
= 22.4 g
Answer: 22.4 g
pKa = -log Ka
3.86 = -log Ka
Ka= 1.38*10^-4
Kb of lactate = 10^-14 / (1.38*10^-4) = 7.244*10^-11
lactate ion + H2O = lactic acid + OH-
1.0 M 0 0
1.0-x x x
Kb = x*x / (1.0-x)
7.244*10^-11 = x^2 /1
x = 8.551*10^-6 M
So,
[OH-] = 8.551*10^-6 M
pOH = -log [OH-]
= -log (8.551*10^-6)
= 5.07
pH = 14 – pOH
= 14 – 5.07
= 8.93
Answer: 8.93