Calculate the percent ionization of formic acid (HCO_2H) in a solution that is 0

Question

Calculate the percent ionization of formic acid (HCO_2H) in a solution that is 0.311 M in formic acid and 0.189 M in sodium formate (NaHCO_2). The K_a of formic acid is 1.77 times 10^-4. 37.8 0.0937 11.3 1.06 times 10^-3 3.529 A 25.0 mL sample of 0.723 M HCLO_4 is titrated with a 0.723 M KOH Solution. What is the [H^+] (morality) at the equivalence point? 0.439 1.00 times 10^-7 0.723 2.81 times 10^-13 0.273 The solubility product of lead chloride is 1.6 times 10^-5. What is the concentration of lead ions in a saturated solution of lead chloride at this temperature? 5.0 times 10^-4 M 4.1 times 10^-6 M 3.1 times 10^-7 M 1.6 times 10^-2 M non of the above

Explanation / Answer

Dear candidate you have asked three questions as per guidelines one question one time, still we solve first two questions. Please send the other one separately.

67. HCO2H <------> H++ HCO2-

Ka = [H+][HCO2-] / [HCO2H] ======1)

NaHCO2 will dissociate completely to form 0.189 M formate ion (HCO2-). The concentration of HCO2- due to formic acid is neglegible.

let x moles /L of formic acid dissociate so that x moles of each of H+ and HCO2- are formed. Hence eqn 3 will be written as

1.77 X 10-4 = x X (0.189) / (0.311 - x)

since x is very small hence can be neglected in denomenator

1.77 X 10-4 = x (0.189) / 0.311

x = 1.77 X 10-4 X 0.311 / 0.189 = 2.91 X 10-4

percent dissociation = ([H+] / [[HCO2H] ) X 100 = (2.91 X 10-4 / 0.311) X 100 = 0.0935%

Hence option (B) is correct answer.

68. Since both acid as well as base are stong hence solution will be neutral at equivalence point, that means H+ concentration will be equal to OH-. For a neutral solution we have

[H+][OH-] = 10-14

hence [H+] = [OH-] = 10-7M

10-7 + 10-7 = 10-14

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