Question

Lactic acid is found in muscles after exercising and has a Ka = 1.4 x 10-4. A 350.0 mL solution contains 0.060 moles of lactic acid and 0.040 moles of sodium lactate. What is the pH of the solution after the addition of 0.015 moles of HNO3?

Answer 1

A Buffer is a solution where both a weak acid and the salt of its conjugate base are present. Now the pH of the buffer is calculated by the formula

pH = pKa + log [salt of conjugate base]/[weak acid]

Now here we have the weak acid = lactic acid

And salt of its conjugate base = sodium lactate

So the pH of our solution is-

pH = pKa + log [sodium lactate]/[lactic acid]

where

pKa = -log Ka

= -log (1.4 x 10^-4)

= 3.85

Now when we add a strong acid like HNO₃, then the H⁺ of the strong acid will react with the lactate ion (sodium lactate -------> sodium⁺ + lactate^- ) to form lactic acid

i.e H⁺ + lactate^- -------------> lactic acid

So due to addition of HNO3, the amount of lactate^- ion from the solution will decrease and that of lactic acid will increase. Now from the equation we can see 1 mole of HNO3 will react with 1 mole of lactate^- to form 1 mole of lactic acid

So the final amount of species present will be-

Reaction	H+ +	lactate- ------------->	lactic acid
Initial	0.015 moles	0.040 moles	0.060 moles
Change	-0.015 moles	-0.015 moles	0.015 moles
Equilibrium	0	0.025 moles	0.075 moles

So putting the values in the formula-

pH = pKa + log [sodium lactate]/[lactic acid]

= 3.85 + log [0.025 moles]/[0.075 moles]

= 3.85 + log [0.33]  

= 3.85 + [-0.47]  

= 3.38‬