Question

What is the pH of a solution that is prepared by dissolving 9.90 grams of pyruvic acid (formula weight = 88.06 grams/mol) and 9.25 grams of sodium pyruvate (formula weight = 90.04 grams/mole) in water and diluting to 800.00 mL? The K_a for pyruvic acid is 0.00324.

Answer 1

Step 1:
mass(HA)= 9.90 g

use:
number of mol of HA,
n = mass of HA/molar mass of HA
=(9.9 g)/(88.06 g/mol)
= 0.1124 mol
volume , V = 8*10^2 mL
= 0.8 L


use:
Molarity,
M = number of mol / volume in L
= 0.1124/0.8
= 0.1405 M

Step 2:
mass(NaA)= 9.25 g

use:
number of mol of NaA,
n = mass of NaA/molar mass of NaA
=(9.25 g)/(90.04 g/mol)
= 0.1027 mol
volume , V = 8*10^2 mL
= 0.8 L


use:
Molarity,
M = number of mol / volume in L
= 0.1027/0.8
= 0.1284 M

Step 3:

Ka = 3.24*10^-3

pKa = - log (Ka)
= - log(3.24*10^-3)
= 2.489

use:
pH = pKa + log {[conjugate base]/[acid]}
= 2.489+ log {0.1284/0.1405}
= 2.45

Answer: 2.45