Buffers- Chemistry

Question

A buffer is prepared by adding 12.0 g lactic acid (MW = 90.1 gmol-1), and 10.0 g sodium lactate (MW = 112 gmol-1), to enough water to form a 500 mL solution. The Ka of lactic acid is 1.38 x 10–4.

a) What is the pH of the buffer?

b) What is the pH of the solution after the addition of 50.0 mL of 0.100 M HCl(aq) to this buffer?

Answer 1

You can use the Henderson Hasselbalch (HH) equation when working with most buffers.

pH = pKa + log [salt]/[acid]

The pKa = -log Ka = -log 1.38x10-4 = 3.86

moles salt = 10.0 g x 1 mol/112 g = 0.0893 moles salt initially present

moles acid = 12.0 g x 1 mol/90.1 g = 0.133 moles acid initially present

[salt] = 10.0 g x 1 mol/112 g /0.5 L = 0.179 M

[acid] = 12.0 g x 1 mol/90.1 g /0.5 L = 0.266 M

pH = pKa + log [salt]/[acid] = 3.86 + log (0.179/0.266)

pH = 3.86 + (-0.17)

pH = 3.69

After adding 50.0 ml of 0.1 M HCl, the final volume will be 500 ml + 50 ml = 550 ml = 0.550 L

The HCl will react with the lactate salt (A-) to produce lactic acid (HA) so [acid] will increase and [salt] will decrease.

Moles HCl added = 0.050 L x 0.1 mol/L = 0.005 moles

Final moles salt = 0.0.0893 mol - 0.005 mol = 0.0843 moles salt

Final moles acid = 0.133 mol + 0.005 mol = 0.138 moles acid

Final [salt] = 0.0843 mol/0.550 L = 0.153 M

Final [acid] = 0.138 mol/0.550 L = 0.251 M

Using the HH equation again, we have...

pH = 3.86 + log (0.153/0.251)

pH = 3.86 + (-0.21)

pH = 3.65

Answer 2

a) In order to find the pH of a buffer system, use the Henderson-Hasselbalch Equation:

pH = pK_a + log ([A^-]/[HA]) where pK_a is the -log(K_a), [HA] is the concentration of the acid part of the conjugate acid/base pair and [A^-] is the concentration of salt that fully dissolves to introduce the conjugate base. You may use moles instead of concentration for the second term, since the solution volume is the same for the acid and conj. base. You obtain the moles by dividing the grams by the molar mass (grams/mole).

The H-H equation is what you get if you take the -log of the K_a expression and solve for pH.

b) You can use the same equation with the assumption that the new n_b' = n_b - n_HCl because the HCl will neutralize the base and create more HA: n_a' = n_a + n_HCl You calculate moles of HCl with Liters x Molarity of HCl added.

Good luck!