What is the concentration of C_2O^2-_4 in a 0.100 M solution of oxalic acid (C_2

Question

What is the concentration of C_2O^2-_4 in a 0.100 M solution of oxalic acid (C_2O_4H_2) in water ? C_2O_4H_2 Equilibrium C_2O_4H^- K_a1 = 5.37 times 10^-2 C_2O_4H^- Equilibrium C_2O^2-_4 + H^+ K_a2 = 5.37 times 10^-2 0.100 M 0.232 M 5.37 times 10^-2 M 7.33 times 10^-3 M 5.37 times 10^-5 M Determine the pH of a buffer made by adding 4.259 g of disodium hydrogen phosphate, Na_2HPO_4, to 100.0 ml of 0.150 M HCl solution. Phosphoric acid (H_3PO_4) has the following acid dissociation constants. K_a1 = 7.1 times 10^-3 K_a2 = 6.3 times 10^-8 K_a3 = 4.5 times 10^-13 5.8 11.5 2.1 12.4 7.2

Explanation / Answer

Apply:

H2A = H+ + HA- Ka1

HA- = H+ + A-2 Ka2

so

Ka1 = (H+)(HA-)/H2A

5.37*10^-2 = x*x/(0.1-x)

x = 0.0511

HC2O4- = 0.0511 M

so

from Ka2:

Ka2 = (H+)(A-2) / (HA-)

5.37*10^-5 = (0.0511+y)*y/(0.0511-y)

y = 0.00163

so

C2O4-2 = 0.00163 = 1.6*10^-3

nearest answer is E

Q18

m = 4.259 g of Na2HPO4

V = 100 mL of 0.15 HCl solution

mol of HPO4- = mass/MW = 4.259/141.96 = 0.03 mol of conjugate = 30 mmol of conjguate

When adding H+

mmol of H2PO4- formed = MV = 0.15*100 = 15 mmol

mmol o fHPO4-2 reacted = 30 - 15 = 15

this is a buffer! since

H2PO4- and HPO4-2 are present

actualyl, this is the second ionization point

so

pKa2 = -log(Ka2) = 7.21

so

pH = pKa2 + log(HPO4-2 / H2PO4-)

pH = 7,21 + log(15/15)

so

pH = 7.21 approx

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