Having trouble filling in the blanks. Please help by showing all work so that I

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Having trouble filling in the blanks. Please help by showing all work so that I may understand how you got to your answer. Thank you in advance!

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

Carbonic acid H2CO3 is a weak acid .

H2CO3 H+ + HCO3-

The Ka1for this is 4.3 × 10-7

The second dissociation ,
HCO3- H+ + CO3 2-

The Ka2 for this is 5.6× 10-11. It is very small value so it can be ignored.

Calculate [H+] in 0.18M solution of H2CO3 using the Ka equation:

Ka 1 = [H+][HCO3-] / [H2CO3]

Here [H+]= [HCO3-]

Ka 1= [H+]² / [H2CO3]

4.3*10^-7 = [H+]² / 0.18

[H+]² = (4.3*10^-7)*( 0.18)

[H+]² = 7.74*10^-8

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

pH = -log [H+]

pH = -log ( 2.78*10^-4)

pH = 3.56

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