Part 1 How many grams of dipotassium phthalate (242.3 g/mol) must be added to 30
ID: 854108 • Letter: P
Question
Part 1
How many grams of dipotassium phthalate (242.3 g/mol) must be added to 300 mL of 0.125 M potassium hydrogen phthalate to give a buffer of pH 5.70? The Ka's for phthalic acid are 1.12 x 10-3 and 3.90 x 10-6.
Part 2
Part 1 How many grams of dipotassium phthalate (242.3 g/mol) must be added to 300 mL of 0.125 M potassium hydrogen phthalate to give a buffer of pH 5.70? The Ka's for phthalic acid are 1.12 x 10 - 3 and 3.90 x 10 - 6. Part 2 Determine the pH at the equivalence point of a titration of phosphoric acid with NaOH. H3PO4 + 2 NaOH Right arrow Na2HPO4+ 2 H2O Assume that the concentration of salt at the equivalence point is 0.03 M.Explanation / Answer
Ka1 = 1.12 x 10^-3; pKa1 = 2.95
Ka2 = 3.90 x 10^-6; pKa2 = 5.41
pH = 5.70 = target pH
[KHP] = 0.125 M
V(KHP) = 300 mL = 0.300 L
molar mass K2P = 243.2 g/mol
Equilibria:
H2P + H2O <=> HP^- + H3O^+ >> Ka1 = [HP^-][H3O^+]/[H2P]
HP^- + H2O <=> P^2- + H3O^+ >> Ka2 = [P^2-][H3O^+]/[HP^-]
You can use the Henderson-Hasselbalch equation to help solve the problem.
pH = pKa2 + log[P^2-]/[HP^-]
5.70 = 5.41 + log[P^2-]/[HP^-]
log[P^2-]/[HP^-] = 0.29
log[P^2-] - log[HP^-] = 0.29
log[P^2-] = 0.29 + log[HP^-] = 0.29 + log(0.125) = -0.61
[P^2-] = 10^(-0.61) = 0.24M
Assume no volume change when adding the solid (good assumption).
moles K2P = V x M = (0.300 L)(0.24 M) = 0.072 mole K2P must be added
mass K2P = (0.072 mole)(242.3 g)/(1 mol) = 17.44 g K2P
Check: pH = 5.41 + log(0.24)/(0.125) = 5.41 + log(2.457) = 5.41 +0.28 = 5.69
2)
pOH = -log[OH-]
ph = 14 - pOH
-log[0.03] = 1.52
14 - 1.52
ph = 12.48
Hope this is helpful to you.