Tabulate the concentration of the species involved in the equilibrium reaction,

Question

Tabulate the concentration of the species involved in the equilibrium reaction, letting x = [H+] at equilibrium (Do not calculate “x” yet)

Use the Ka expression and determine the concentration of H+ and the pH of this solution.

First, fill out the BCA table to determine the moles of material at this point in the titration. (Fill in actual values, do not use x. Also be sure to use moles)

The resulting concentrations of CH3COOH and CH3COO- after the reaction here are:

Use these calculated concentrations and plug into the following ICE table. Let x = [H+] at equilibrium (Do not calculate “x” yet)

Use the Ka expression and determine the concentration of H+ and the pH of this solution.

What volume of the titrant has been added to the starting material at the equivalence point for this titration?

mL

At the equivalence point for a weak acid-strong base titration an equal number of moles of OH- and H+have reacted, producing a solution of water and salt. What affects the pH at this point for a strong-acid/strong-base titration?

First, fill out the BCA table to determine the moles of material at the equivalence point of the titration. (Fill in actual values, do not use x. Also be sure to use moles)

Use these calculated concentrations and plug into the following ice table. Let x = [OH-] at equilibrium (Do not calculate “x” yet)

Use the Ka expression and determine the concentration of OH- and the pOH of this solution.

For all of the following questions 10.00mL of 0.187 M acetic acid (CH3COOH) is titrated with 0.100M KOH
(The Ka of acetic acid is 1.80 x 10-5)
Region 1: Initial pH
Before any titrant is added to the starting material

Tabulate the concentration of the species involved in the equilibrium reaction, letting x = [H+] at equilibrium (Do not calculate “x” yet)

CH3COOH H+(aq) CH3COO-(aq) Initial concentration (M) Change in
concentration (M) -x +x +x Equilibrium
concentration (M)

Use the Ka expression and determine the concentration of H+ and the pH of this solution.

[H+] = x 10  M pH = Region 2: Before the Equivalence Point
12.83mL of the 0.100M KOH has been added to the starting material

First, fill out the BCA table to determine the moles of material at this point in the titration. (Fill in actual values, do not use x. Also be sure to use moles)

CH3COOH KOH H2O(l) KCH3COO(aq) B X C X A X

The resulting concentrations of CH3COOH and CH3COO- after the reaction here are:

[CH3COOH] = M [CH3COO-] = M

Use these calculated concentrations and plug into the following ICE table. Let x = [H+] at equilibrium (Do not calculate “x” yet)

CH3COOH H+(aq) CH3COO-(aq) Initial concentration (M) Change in
concentration (M) -x +x +x Equilibrium
concentration (M)

Use the Ka expression and determine the concentration of H+ and the pH of this solution.

[H+] = x 10  M pH = Region 3: At the Equivalence Point

What volume of the titrant has been added to the starting material at the equivalence point for this titration?

mL

At the equivalence point for a weak acid-strong base titration an equal number of moles of OH- and H+have reacted, producing a solution of water and salt. What affects the pH at this point for a strong-acid/strong-base titration?

None of these The auto-ionization of water The basicity of the salt anion The acidity of the salt cation

First, fill out the BCA table to determine the moles of material at the equivalence point of the titration. (Fill in actual values, do not use x. Also be sure to use moles)

CH3COOH KOH H2(l) KCH3COO(aq) B X C X A X

Use these calculated concentrations and plug into the following ice table. Let x = [OH-] at equilibrium (Do not calculate “x” yet)

CH3COO- H2O(l) OH-(aq) CH3COOH(aq) Initial concentration (M) X Change in
concentration (M) X Equilibrium
concentration (M) X

Use the Ka expression and determine the concentration of OH- and the pOH of this solution.

[OH-] = x 10  M pOH =

Explanation / Answer

Question 1.

Region 1: Initial pH

First, assume the acid:

CH3COOH

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = 0.187 ;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = 0.187 - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 0.187 M; then

x^2 + (1.8*10^-5)x - 0.187 *(1.8*10^-5) = 0

solve for x

x =0.001825

substitute

[H+] = 0 + 0.001825= 0.001825M

[A-] = 0 + 0.001825= 0.001825M

[HA] = M - x = 0.187 -0.001825= 0.185175 M

pH = -log(H+) = -log(0.001825) = 2.738

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