Concentrations of Solutions Titrations Weak Acids and Bases

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (e) 35.5 mL, (f) 50.0 mL.

The initial pH of 35.0 mL 0.150 M CH3COOH is 2.784. After 17.5 mL NaOH added, the pH is 4.74. After 34.5 mL NaOH added, the pH is 6.584. After 35.0 mL NaOH added, the pH is 8.810. After 35.5 mL NaOH added, the pH is 9.88. After 50.0 mL NaOH added, the pH is 11.35.

Weak Acid Equilibrium Calculations

Acetic acid has a Ka value of 1.8E-5. This will be very important to use when calculating the pH of the initial solution as well as the solutions after the specified amounts of base have been added.

(a) Initial pH of 35.0mL 0.150 M CH3COOH

To calculate the initial pH of the weak acid solution (when 0 mL NaOH have been added), the initial concentration of acetic acid and the acid-ionization constant is used to find the amount of hydrogen ions at equilibrium.

[H+][CH3COO-] =Ka [CH3COOH]

CH3COOH <–> CH3COO- + H+

(x)(x)/(0.150-x) =1.8E-5

assume x << 5%

(x^2)/(0.150) =1.8E-5

x=[H+]=1.64E-3 M

CH3COOH <–> CH3COO- + H+
i0.150 M0 M0 M
c-x+x+x
e0.150 – x x x

To make sure the assumption of percent ionization is less than 5% we divide the change, x, by the initial molarity of weak acid, then multiply the value by 100:

assumption for 5% rule of ionization to eliminate the need to use the quadratic formula

The assumption is valid meaning the change in x value calculated can be used to find the pH of the initial 35.0 mL 0.150 M acetic acid solution:

x=[H+]=1.64E-3 M

pH = -log(1.64E-3) = 2.784

Mole Calculations for the Titration

It is important to first calculate how many moles of acetic acid are present before any additions of sodium hydroxide:

Converting 35.0 mL acetic acid of 0.150 M acetic acid to moles of acetic acid

0.00525 moles of acetic acid are initially present in the 35.0 mL 0.150 M acetic acid solution.

pH Calculations Before the Equivalence Point

(b) pH after 17.5 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

Acetic acid reacting with hydroxide ions in the buffer region of titration before the equivalence point.

CH3COOH + OH- —-> CH3COO- + H2O

There are 0.00525 moles of acetic acid initially present in 35.0 mL 0.150 M acetic acid solution. When 17.5 mL 0.150 M NaOH are added, 0.002625 moles of hydroxide ions are added to the weak acid solution and react with acetic acid:

(moles)CH3COOH–>CH3COO-+H2O
before addition0.00525 mol0 molx
addition OH--0.002625 mol+0.002625 molx
after addition0.002625 mol0.002625 mol

After addition of 17.5 mL 0.150 M NaOH, all hydroxide ions have reacted to produce 0.002625 moles of acetate anion, leaving 0.002625 moles of acetic acid unreacted in solution. The new volume of the solution will equal the sum of the initial volume of weak acid and the volume of sodium hydroxide added. With a new total volume, the concentration values of acetic acid and acetate anion will equal the moles after addition divided by the new total volume of solution.

Total Volume = 35.0 mL + 17.5 mL = 52.5 mL solution

[CH3COOH] = [CH3COO-] = 0.002625 mol/0.0525 L = 0.05 M

[H+][CH3COO-] =Ka [CH3COOH]

CH3COOH <–> CH3COO- + H+

(x)(0.05+x)/(0.05-x) =1.8E-5

assume x << 5%

(0.05x)/(0.05) =1.8E-5

x=[H+]=1.8E-5 M

CH3COOH <–> CH3COO- + H+
i0.05 M0.05 M0 M
c-x+x+x
e0.05 – x 0.05 + x x

1.8E-5/0.05 x 100 << 5% meaning the assumption is valid.

x=[H+]=1.8E-5 M

pH = -log(1.8E-5) = 4.74

(c) pH after 34.5 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

Acetic acid reacting with hydroxide ions before the equivalence point.

There are 0.00525 moles of acetic acid initially present in 35.0 mL 0.150 M acetic acid solution. When 34.5 mL 0.150 M NaOH are added, 0.005175 moles of hydroxide ions are added to the weak acid solution and react with acetic acid:

(moles)CH3COOH–>CH3COO-+H2O
before addition0.00525 mol0 molx
addition OH--0.005175 mol+0.005175 molx
after addition7.5E-5 mol0.005175 mol

After addition of 34.5 mL 0.150 M NaOH, all hydroxide ions have reacted to produce 0.005175 moles of acetate anion, leaving 7.5E-5 moles of acetic acid unreacted in solution. The new volume of the solution will equal the sum of the initial volume of weak acid and the volume of sodium hydroxide added. With a new total volume, the concentration values of acetic acid and acetate anion will equal the moles after addition divided by the new total volume of solution.

Total Volume = 35.0 mL + 34.5 mL = 69.5 mL solution

[CH3COO-] = 0.005175 mol/0.0695 L = 0.0745 M

[CH3COOH] = 7.5E-5 mol/0.0695 L = 1.08E-3 M

[H+][CH3COO-] =Ka [CH3COOH]

CH3COOH <–> CH3COO- + H+

(x)(0.0745 +x)/(1.08E-3-x) =1.8E-5

assume x << 5%

(0.0745x)/(1.08E-3) =1.8E-5

x=[H+]=2.61E-7 M

CH3COOH <–> CH3COO- + H+
i1.08E-3 M0.0745 M0 M
c-x+x+x
e1.08E-3 – x 0.0745 + x x

2.61E-7/1.08E-3 x 100 << 5% meaning the assumption is valid.

x=[H+]=2.61E-7 M

pH = -log(2.61E-7) = 6.584

pH Calculations at the Equivalence Point

(d) pH after 35.0 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

Acetic acid reacting with hydroxide ions at the equivalence point.

There are 0.00525 moles of acetic acid initially present in 35.0 mL 0.150 M acetic acid solution. When 35.0 mL 0.150 M NaOH are added, 0.00525 moles of hydroxide ions are added to the weak acid solution and react completely with all moles of acetic acid present:

(moles)CH3COOH–>CH3COO-+H2O
before addition0.00525 mol0 molx
addition OH--0.00525 mol+0.00525 molx
after addition0 mol0.00525 mol

After addition of 35.0 mL 0.150 M NaOH, all hydroxide ions have reacted to produce 0.00525 moles of acetate anion, leaving 0 moles of acetic acid unreacted in solution. The new volume of the solution will equal the sum of the initial volume of weak acid and the volume of sodium hydroxide added. With a new total volume, the concentration value for acetate anion will equal the moles after addition divided by the new total volume of solution.

Total Volume = 35.0 mL + 35.0 mL = 70.0 mL solution

[CH3COO-] = 0.00525 mol/0.0700 L = 0.0750 M

CH3COO- + H2O <–> CH3COOH + OH

[OH-][CH3COOH] =Kb [CH3COO-]

(Ka)(Kb) = 1E-14

Kb = (1E-14)/ (1.8E-5) = 5.6E-10

(x)(x)/(0.0750-x) =5.6E-10

assume x << 5%

(x^2)/(0.0750) =5.6E-10

x=[OH-]=6.45E-6 M

CH3COO- <–> CH3COOH +OH-
i0.0750 M0 M0 M
c-x+x+x
e0.0750 – x x x

6.45E-6/0.0750 x 100 << 5% meaning the assumption is valid.

x=[OH-]=6.45E-6 M

pOH = -log(6.45E-6) = 5.190

pH = 14 – pOH = 14 – 5.190 = 8.810

pH Calculations After the Equivalence Point

(e) pH after 35.5 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

After the equivalence point, any additional volumes of sodium hydroxide added will be solely used to calculate the pH of the solution. Since 35.0 mL NaOH 0.150 M NaOH was needed to react completely with 35.0 mL 0.150 M CH3COOH, the volume used to calculate the moles of hydroxide ions present will be the difference between the volume of sodium hydroxide added and the volume of base at the equivalence point.

Excess Volume NaOH = 35.5 mL – 35.0 mL = 0.5 mL excess 0.150 M NaOH

pOH = -log(7.5E-5) = 4.12

pH = 14 – 4.12 = 9.88

(f) pH after 50.0 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

After the equivalence point, any additional volumes of sodium hydroxide added will be solely used to calculate the pH of the solution. Since 50.0 mL NaOH 0.150 M NaOH was needed to react completely with 35.0 mL 0.150 M CH3COOH, the volume used to calculate the moles of hydroxide ions present will be the difference between the volume of sodium hydroxide added and the volume of base at the equivalence point.

Excess Volume NaOH = 50.0 mL – 35.0 mL = 15.0 mL excess 0.150 M NaOH

pOH = -log(2.25E-3) = 2.648

pH = 14 – 2.648 = 11.35

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