A 100.0 mL sample of 0.18 H2CO3 is titrated with 0.10 M NaOH. Fill in the follow

Question

A 100.0 mL sample of 0.18 H2CO3 is titrated with 0.10 M NaOH. Fill in the following information in the table provided. Carbonic acid is a diprotic acid whose Ka1 = 4.3*10^-7 and Ka2=5.6*10^-11. Please show all work so that I may understand how you reached your answer & to better understand the process.

For calculations please use tions please use cerrect units and sgnilicant llgures. 12C03 is titrated with 0.10 M NaOH. Fill in the following I) A 100.0 mL sample of 0.18 M H information in the table provided. Please show calculations below table. Carbonic acid is a diprotic acid whose Kal = 4.3 × 10-7 and Ka2 = 5.6 x 10-11 . (1 point per blank; 14 pts) pH-relevant chemical species present Titration point Volume of NaOH added pH of solution 0.0 mL H2C03 Initial conditions Equivalence point l Equivalence point 2 H2C03 and HCO3 ½ Equivalence point l ½ Equivalence point 2 HCO3 and CO3 40 mL of NaOH added 40.0 mL H2C03 and HCO3 500.0 ml 500 mL of NaOH added -0.10M NaOH hal-4.3x167

Explanation / Answer

Titration

(1) Initial pH

H2CO3 <==> H+ + HCO3-

let x amount has dissociated

Ka1 = [H+][HCO3-]/[H2CO3]

4.3 x 10^-7 = x^2/0.18

x = [H+] = 2.78 x 10^-4 M

pH = -log[H+] = 3.55

(2) First half Eq point

[H2CO3] = [HCO3-]

pH = pKa1 + log([HCO3-]/[H2CO3])

      = 6.36

(3) First Eq point

[H2CO3] = [NaOH]

Volume of NaOH added = 0.18 M x 100 ml/0.1 M = 180 ml

pH = 1/2(pKa1 + pKa2) = 1/2(6.36 + 10.25) = 8.30

(4) Second half Eq point

[HCO3-] = [CO3^2-]

pH = pKa2 + log([CO3^2-]/[HCO3-])

     = 10.25

(5) Second Eq point

[NaOH] = 2 x [H2CO3]

[CO3^2-] formed = 0.18 M x 100 ml/2 x 180 ml = 0.05 M

CO3^2- + H2O <==> HCO3- + OH-

let x amount has reacted

Kb2 = Kw/Ka2 = [HCO3-][OH-]/[CO3^2-]

1 x 10^-14/5.6 x 10^-11 = x^2/0.05

x = [OH-] = 2.99 x 10^-3 M

pOH = -log[OH-] = 2.52

pH = 14 - pOH = 11.48

(6) 40 ml of 0.1 M NaOH added

moles of H2CO3 = 0.18 M x 100 ml = 18 mmol

moles of NaOH = 0.1 M x 40 ml = 4 mmol

[HCO3-] formed = 4 mmol/140 ml = 0.03 M

[H2CO3] remins = 14 mmol/140 ml = 0.1 M

pH = pKa1 + log(base/acid)

     = 6.36 + log(0.03/0.1)

     = 5.84

(7) 500 ml NaOH added

excess [NaOH] = 0.1 M x (500-360)ml/600 ml = 0.023 M

pOH = -log[OH-] = -log(0.023) = 1.64

pH = 14 - pOH = 12.36

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