How many grams of sodium lactate (MW = 112 g/mol) are required to prepare 200 ml...

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How many grams of sodium lactate (MW = 112 g/mol) are requiredto prepare 200 ml solution at 1.0 M? Given pKa of lactic acid =3.86, what is the pH of this solution?

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3.9 Ratings (467 Votes)

number of moles of sodium Lactate = M*V

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1M * 0.2 L

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.2 mol

mass of sodium lactate = MW * number of moles

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 112 g/mol * 0.2 mol

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 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 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0 Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â

1.0-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â x

Kb = x*x / (1.0-x)

7.244*10^-11 Â Â = x^2 /1

x = 8.551*10^-6 M

So,

[OH-] = 8.551*10^-6 M

pOH = -log [OH-]

Â Â Â Â Â = -log (8.551*10^-6)

Â Â Â Â Â = 5.07

pH = 14 â€“ pOH

Â Â Â = 14 â€“ 5.07

Â Â Â = 8.93

Answer: 8.93

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