Question

a. How many grams of sodium lactate, NaC3H5O3 should be added to 1.00L of 0.150M lactic acid, HC3H5O3 to form a buffer solution with a pH of 3.90? (Ka= 1.4 x 10^-4 for lactic acid)

b. What is the pH of the buffer solution containing 0.10moles of acetic acid and 0.13moles of sodium acetate in 1.0L after adding 0.02 moles of KOH?

c. What will the pH be if 0.15mol HA (pKa= 2.00) and 0.27 mole A- are mixed together to make a 2.00L solution?

Answer #1

Sol :-

K_{a} = 1.4 x 10^{-4}

pK_{a} = - log K_{a} = - log(1.4
x10^{-4}) = 3.8539

**Using Henderson-Hasselbalch equation, we
have**

pH = pK_{a} + log
([NaC_{3}H_{5}O_{3}]/[HC_{3}H_{5}O_{3}])

3.9 = 3.8539 + log
([NaC_{3}H_{5}O_{3}]/[HC_{3}H_{5}O_{3}])

log
([NaC_{3}H_{5}O_{3}]/[HC_{3}H_{5}O_{3}])
= 0.0461

[NaC_{3}H_{5}O_{3}]/[HC_{3}H_{5}O_{3}]=
1.1121

[NaC_{3}H_{5}O_{3}] = 0.1668

volume , V = 1.0 L

We know ,

number of mol = Molarity x Volume

= 0.1668 x 1

= 0.1668 mol

Molar mass of NaC_{3}H_{5}O_{3}
= 112.06 g/mol

mass of NaC_{3}H_{5}O_{3} =
number of mol x gram molar mass

= 0.1668 mol * 112.06 g/mol

= 18.7 g

**Hence, mass of NaC _{3}H_{5}O_{3}
required = 18.7 g**

1. How many grams of sodium lactate, NaC3H5O3, should be
added to 1.00 L of 0.150 M lactic acid, HC3H5O3, to form a buffer
solution with a pH of 3.90? (Ka = 1.4*10^-4 for lactic acid)
How do you do this?

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(c) Plot a graph showing a mole fraction of Lactic acid and
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