Question

In: Chemistry

In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of...

In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of the resultant solution after the addition of 25.00 mL of 0.100 M KOH? Assume that the volumes of the solutions are additive. Ka = 1.8x10-5 for CH3COOH.

Solutions

Expert Solution

The balanced chemical equation for your neutralization reaction was

CH3COOH(aq)+OH−(aq)→CH3COO−(aq)+H2O(l)

Calculate how many moles of acetic acid and of potassium hydroxide you add together

C=nV⇒n=C⋅V

nCH3COOH=0.100 M⋅25⋅10−3L=0.0025 moles

nNaOH=0.100 M⋅25⋅10−3L=0.0025 moles

Since you add equal numbers of moles of acetic acid and potassium hydroxide, bothcompounds will be completely consumed by the reaction.

In the process, the reaction will produce 0.0025 moles of acetate anions, CH3COO−.

nCH3COOH=0.0025−0.0025=0

nKOH=0.0025−0.0025=0

nCH3COO−=0.0025 moles

The acetate anion can act as a weak base and react with water to produce hydroxide ions and reform acetic acid according to the following equilibrium

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)

To determine the concentration of the acetate anion, use the total volume of the solution, which is

Vsol=Vacetic acid+Vpotassium hydroxide

Vsol=25+25=50. mL

This will get you

[CH3COO−]=0.0025 moles50.⋅10−3L=0.0500 M

Use an ICE table to determine the concentration of the hydroxide ions in solution

CH3COO−(aq)+H2O(l)⇌CH3COOH(aq)+OH−(aq)
I.......0.0500.............................................0.............................0
C........(-x).................................................(+x)..........................(+x)
E......0.0500-x..........................................x..............................x

By definition, the base dissociation constant, Kb, will be (ka.kb= 1x 10 -14) then Kb=5.6X10−10

Kb=[OH−]⋅[CH3COOH]/ [CH3COO−]

= x⋅x/0.0500−x= x2/0.0500−x

Since Kb is so small, you can approximate (0.0500 - x) with 0.0500. This will get you

Kb = x2 /0.0500=5.6⋅10−10⇒ x= 5.29⋅10−6

Since x=[OH−], you have

[OH−]=5.29⋅10−6M

The pOH of the solution will be

pOH=−log([OH−])=−log(5.29⋅10−6)=5.28

Therefore, the pH of the solution will be

pHsol=14−pOH=14−5.28=8.72


Related Solutions

In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of...
In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of the resultant solution after the addition of 15.06 mL of 0.100 M of KOH? Assume that the volumes of the solutions are additive. Ka = 1.8x10-5 for acetic acid, CH3COOH.
What is the pH at the equivalence point for the titration of 25.00 mL 0.100 M...
What is the pH at the equivalence point for the titration of 25.00 mL 0.100 M CH3CH2CH2CO2- with 0.1848 M HCl? Species (K values) CH3CH2CH2CO2H (Ka = 1.14E-5) CH3CH2CH2CO2- (Kb = 2.63x10-10
Consider the titration of 25.0 mL of 0.100 M acetic acid (HA) with 0.100 M NaOH....
Consider the titration of 25.0 mL of 0.100 M acetic acid (HA) with 0.100 M NaOH. 1. Write the balanced chemical equation and equilibrium constant expression (ECE) for all of the reactions that occur when NaOH is added to the acetic throughout the titration. Hint: think of what is in the solution (acetic acid) with water, acetic acid with sodium hydroxide, and acetate ion with water) as the titration is proceeding. 2. Calculate the volume of NaOH solution needed to...
For the titration of 50.0 mL of 0.150 M acetic acid with 0.100 M sodium hydroxide,...
For the titration of 50.0 mL of 0.150 M acetic acid with 0.100 M sodium hydroxide, determine the pH when: (a) 50.0 mL of base has been added. (b) 75.0 mL of base has been added. (c) 100.0 mL of base has been added.
titration of 25.00 ml of 0.100M CH3COOH with 0.100 M NaOH (weak acid, strong base) a)...
titration of 25.00 ml of 0.100M CH3COOH with 0.100 M NaOH (weak acid, strong base) a) calculate the initial pH ( kb = 1.8 x10 ^-5) b) why is pH > 7 at the equivalence point? c) calculate the pH at the equivalence point
25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH...
25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH. Calculate the pH of the solution after addition of 50 mL of NaOH. Ka(CH3COOH) = 1.8 x 10-5.
Calculate the pH of a titration of 50.00 mL of 0.100 M Phenylacetic acid , Ka...
Calculate the pH of a titration of 50.00 mL of 0.100 M Phenylacetic acid , Ka = 4.9 x 10-5, with 0.100 M NaOH at the following points: SHOW ALL WORK IN NEAT DETAIL ON A SEPARATE PAGE. (Be sure to write chemical equations and Ka or Kb expressions when needed.) a. (4 Pts) Before any NaOH is added. b. (4 Pts) After 18.7 mL of NaOH are added. c. (4 Pts) After 25.00 mL of NaOH are added. d....
25.00 mL 0.100 M CH3NH2 is titrated with 0.100 M HCl. Calculate the pH of this...
25.00 mL 0.100 M CH3NH2 is titrated with 0.100 M HCl. Calculate the pH of this the titration solution after the addition of a) 12.5 mL and b) 25.00 mL of the titrant has been added. Based on these pH values, select an appropriate indicator for the titration from table at the end of lecture notes for Chapter 16.
Calculate the pH of the titration solution or 25.00 mL of 0.255 M nitrous acid nitrates...
Calculate the pH of the titration solution or 25.00 mL of 0.255 M nitrous acid nitrates with 0.214 M KOH at the following volumes of KOH: A. 0.00 mL B. 10.00 mL C. Mid-point of titration D. Equivalance point of titration E. 5.00 mL past equivalence point of titration
Assume a titration with 0.100 M NaOH titrant and 25.00 mL of a 0.0800 M CH3COOH...
Assume a titration with 0.100 M NaOH titrant and 25.00 mL of a 0.0800 M CH3COOH analyte. What will the pH of the analyte be if 10.00 mL of NaOH is added?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT