CHM 107, Quiz 2 Name 1. going to formic acid with strong base, is 100. to of 0.5

Question

CHM 107, Quiz 2 Name 1. going to formic acid with strong base, is 100. to of 0.50 M formic acid and the concentration of NaoH is 1.0 M. All work must be shown receive credit. The quiz will be due at the beginning of class on 4/24/17. 1) What is the initial pH of the formic acid solution? 2) What is the percent ionization under initial conditions? 3) After the addition of 10 mL ofNaOH, what is the pH? 4) After the addition of 25 mL of NaOH, what is the pH? Think about where in the titration this brings you.

Explanation / Answer

HCOOH (aq) <----> HCOO- + H+
      
1) what is the initial pH of the formic acid solution:

       [HCOOH](M)                 [H+](M)           [HCOO-](M)

Initial                0.50                             0.00                 0.00

Change                    -x                               +x                    +x

Equilibrium          0.50 -x                               x                      x

Ka = [HCOO-][H+]/[HCOOH]

1.7 x 10^-4 = (x)(x)/ (0.50 - x) {x is very small so, 0.5 - x ~ x}

1.7 x 10-4 = x^2/0.5
x^2 = 8.5 x 10^-5
x = 9.22 x 10-3

therefore [H+] = [HCOO-] = 9.22 x 10^-3M

[HCOOH] = 0.500 - 0.00922 ~ 0.490M

2) percent of ionization, this will represent the relative number of acid molecules
which dissociate. It is calculated as :

percent ionization = {[H+]eq/ [HCOOH]i} x 100 = {(9.22 x 10^-3M)/(0.5 M)} x 100
= 1.84 %

3) after addition of 10 mL of 1M NaOH ;

No. moles of HCOOH = 0.1 L x 0.5 M = 0.05 moles
No. moles of NaOH = 0.01 L x 1.0 M = 0.01 moles

Moles of HCOOH = 0.05 - 0.01 moles = 0.04 moles / 0.11 L
therefore, molarity of [HCOOH] = 0.364 M

Molarity of HCOO- = [HCOO-] = 0.01 moles/ 0.11L = 0.091 M

we can use Henderson–Hasselbalch equation ;

pH = pKa + log [base]/[Acid]
pH = 3.77 + log [HCOO-]/[HCOOH]

pH = 3.77 + log 0.091/0.344 = 4.35

4) After addition of 25 mL of NaOH
No. moles of HCOOH = 0.1 L x 0.5 M = 0.05 moles
No. moles of NaOH = 0.025 x 1.0 M = 0.025moles

Moles of HCOOH = 0.05 - 0.025 moles = 0.025 moles / 0.125 L
therefore, molarity of [HCOOH] = 0.2 M

Molarity of HCOO- = [HCOO-] = 0.025 moles/ 0.125L = 0.2 M

we can use Henderson–Hasselbalch equation ;

pH = pKa + log [base]/[Acid]
pH = 3.77 + log [HCOO-]/[HCOOH]

pH = 3.77 + log (0.2/0.2)
pH = 3.77

This is the half ionization where pH = pKa = 3.77

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