Question

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?

Answer 1

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