Question

2 2) WA 4.8 22) Consider the titration of a 20.0-ml sample of 0.105 M HC2H3O2 with 0.125 M NaOH. 3 Determine the pH at the equivalence point. 2 pka 0.125 0.00zł moi A) 2.86 B) 4.74 x=onellt= 16.8 ml C) 8.75 D) 12.17 0.105MOL = 0.0021 mo
please show work

Answer 1

find the volume of NaOH used to reach equivalence point
M(HC2H3O2)*V(HC2H3O2) =M(NaOH)*V(NaOH)
0.105 M *20.0 mL = 0.125M *V(NaOH)
V(NaOH) = 16.8 mL
Given:
M(HC2H3O2) = 0.105 M
V(HC2H3O2) = 20 mL
M(NaOH) = 0.125 M
V(NaOH) = 16.8 mL


mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)
mol(HC2H3O2) = 0.105 M * 20 mL = 2.1 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.125 M * 16.8 mL = 2.1 mmol


We have:
mol(HC2H3O2) = 2.1 mmol
mol(NaOH) = 2.1 mmol

2.1 mmol of both will react to form C2H3O2- and H2O

C2H3O2- here is strong base
C2H3O2- formed = 2.1 mmol
Volume of Solution = 20 + 16.8 = 36.8 mL
Kb of C2H3O2- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10
concentration ofC2H3O2-,c = 2.1 mmol/36.8 mL = 0.0571M

C2H3O2- dissociates as

C2H3O2-        + H2O   ----->     HC2H3O2 +   OH-
0.0571                        0         0
0.0571-x                      x         x


Kb = [HC2H3O2][OH-]/[C2H3O2-]
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 ((5.556*10^-10)*5.707*10^-2) = 5.631*10^-6

since c is much greater than x, our assumption is correct
so, x = 5.631*10^-6 M



[OH-] = x = 5.631*10^-6 M

use:
pOH = -log [OH-]
= -log (5.631*10^-6)
= 5.2495


use:
PH = 14 - pOH
= 14 - 5.2495
= 8.7505
Answer: C