# A chemist titrates 190.0 mL of a 0.4427 Macetic acid (HCH,CO, solution with 0.2382 M NaOH...

A chemist titrates 190.0 mL of a 0.4427 Macetic acid (HCH,CO, solution with 0.2382 M NaOH solution at 25 "C. Calculate the pH at equivalence. The pk of acetic acid is 4.76. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added. pH = 0 x 5 ?

use:

pKa = -log Ka

4.76 = -log Ka

Ka = 1.738*10^-5

find the volume of NaOH used to reach equivalence point

M(HCH3CO2)*V(HCH3CO2) =M(NaOH)*V(NaOH)

0.4427 M *190.0 mL = 0.2382M *V(NaOH)

V(NaOH) = 353.1192 mL

Given:

M(HCH3CO2) = 0.4427 M

V(HCH3CO2) = 190 mL

M(NaOH) = 0.2382 M

V(NaOH) = 353.1192 mL

mol(HCH3CO2) = M(HCH3CO2) * V(HCH3CO2)

mol(HCH3CO2) = 0.4427 M * 190 mL = 84.113 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2382 M * 353.1192 mL = 84.113 mmol

We have:

mol(HCH3CO2) = 84.113 mmol

mol(NaOH) = 84.113 mmol

84.113 mmol of both will react to form CH3CO2- and H2O

CH3CO2- here is strong base

CH3CO2- formed = 84.113 mmol

Volume of Solution = 190 + 353.1192 = 543.1192 mL

Kb of CH3CO2- = Kw/Ka = 1*10^-14/1.738*10^-5 = 5.754*10^-10

concentration ofCH3CO2-,c = 84.113 mmol/543.1192 mL = 0.1549M

CH3CO2- dissociates as

CH3CO2- + H2O -----> HCH3CO2 + OH-

0.1549 0 0

0.1549-x x x

Kb = [HCH3CO2][OH-]/[CH3CO2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.754*10^-10)*0.1549) = 9.44*10^-6

since c is much greater than x, our assumption is correct

so, x = 9.44*10^-6 M

[OH-] = x = 9.44*10^-6 M

use:

pOH = -log [OH-]

= -log (9.44*10^-6)

= 5.025

use:

PH = 14 - pOH

= 14 - 5.025

= 8.975

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