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 ([NaC₃H₅O₃]/[HC₃H₅O₃])

3.9 = 3.8539 + log ([NaC₃H₅O₃]/[HC₃H₅O₃])

log ([NaC₃H₅O₃]/[HC₃H₅O₃]) = 0.0461

[NaC₃H₅O₃]/[HC₃H₅O₃]= 1.1121

[NaC₃H₅O₃] = 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 NaC₃H₅O₃ =  112.06 g/mol

mass of NaC₃H₅O₃   = number of mol x gram molar mass

= 0.1668 mol * 112.06 g/mol

= 18.7 g

Hence, mass of NaC₃H₅O₃ required = 18.7 g