Question 3 5 poitts A 250 mL solution of 010 M acetic acid K

Question 3 (5 poitts) A 25.0 mL solution of 0.10 M acetic acid (K 1.8 x 10-5) is titrated with a 0.1 M NaOH solution. The pH at equivalence is: srthe ) O a) pH = 8.72 O b) pH 5.60 O c) pH- 9.13 O d) pH 5.29 Save Question 4 (5 points) A 25.0 mL of a weak acid is titrated with a strong base (0.1 M). Calculate the pH of the solution during the titration if the weak acid concentration is 0.10 M and its K 1.8 x 10 and 10.0 mL of base has been added. (Hint: use Henderson-Hasselbach equation). O a) pH = 4.56 b) pH- 5.28 O c) pH= 7.00 O d) pH= 4.74

Solution

3.

acetic acid = weak acid

NaOH = strong base

this is a weak acid strong base titration.

At the equivalence point all the millimoles of acid have been consumed. First calculate # of millimoles of acetic acid = volume * molarity

25 mL * 0.10 M = 2.5 millimoles

so in order for 2.5 millimoles of Acetic Acid to be consumed 2.5 millimoles of NaOH must be added.

Since they are both at the same concentration that means the volume has to be the same.

we have added 25 mL of NaOH to reach the equivalence point (total volume = 50 mL)

Let\'s write an equation that determines this
HC2H3O2    +    OH- <=====>    C2H3O2- +    H2O

So, from this you can see that only acetate (C2H3O2-) is present at the equivalence point because all the Acetic acid and NaOH have been consumed. 2.5 millimoles of acetate is formed.

So, [C2H3O2-] = 2.5 milimoles / 50 mL = 0.05 M

However

C2H3O2- will reach an equilibrium.

C2H3O2- + H2O    <========> HC2H3O2    +    OH-
initial: 0.05    0 0
change: -x    +x +x
final: 0.05 - x    x x

now we are at equilibrium so let\'s write an equilibrium problem

Kb (acetate is a base) = x² / ( 0.05 - x)

Ka = 1.8 x 10^-5

Kb = Kw/Ka
= (1.0 x 10^-14) / (1.8 x 10^-5)
= 5.6 x 10^-10

So,
Kb = x² / ( 0.05 - x)
or, 5.6 x 10^-10 = x² / ( 0.05 - x)
or, 5.6 x 10^-10 = x² / ( 0.05) Since Kb is very small, so, we can ignore x in the denominator
or, x² = 2.8 x 10^-11
or, x² = 5.3 x 10^-6

So, x = [OH-] = 5.3 x 10^-6

Then pOH = - log [OH-]
   = - log (5.3 x 10^-6)
   = 5.28

pH = 14 - pOH
   = 14 - 5.28
= 8.72

Answer is option (a)

4.

Let the weak acid be HA and strong base be B.

HA + B    <------> A-    + BH

Moles of HA = 0.1 M x 0.025 L = 0.0025 moles
Moles of B = 0.1 M x 0.010 L = 0.0010 moles

So, 0.0010 moles of B will react with 0.0010 moles of HA to produce 0.0010 moles of A-.
Moles of HA left = 0.0025 moles - 0.0010 moles = 0.0015 moles

Total volume = 25 mL + 10 mL = 35 mL = 0.035 L

So, the concentration terms are

[A-] = 0.0010 moles / 0.035 L = 0.029
[HA] = 0.0015 moles / 0.035 L = 0.043

Ka = 1.8 x 10^-5

pKa = - log Ka
= - log (1.8 x 10^-5)
   = 4.74

Using henderson-hesselbalach equation, we get

pH = pKa + log { [salt] / [acid] }
= pKa + log { [A-] / [HA] }
   = 4.74 + log { 0.029 / 0.043 }
     = 4.74 + log { 0.674 }
     = 4.74 - 0.17
     = 4.57

Answer is option (A) 4.56