Question

Learning Goal:

To learn about titration types and how to calculate pH at different points of titration.

In an acid-base titration, a titrant (solution of a base or acid) is added slowly to an analyte (solution of an acid or base). The titration is often monitored using a pH meter. A plot of pH as a function of the volume of titrant added is called a pH titration curve.

Prior to the titration, the pH is determined by the concentration of the analyte. When the titrant is added, it begins to neutralize the analyte. The decrease in the analyte concentration changes the pH. At the equivalence point, equivalent quantities of acid and base have been mixed together such that the acid-base reaction is complete, and the pH is determined by the product. At 25∘C: in the case of a strong acid-strong base titration, the salt formed is neutral and pH=7; in a weak acid-strong base titration, a basic salt is produced and pH>7; and in a weak base-strong acid titration, the salt is acidic and pH<7. After the equivalence point, the pH is determined by the concentration of excess titrant.

Note: For simplicity and clarity, assume thoughout this item that the temperature is 25∘C.

Part C

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 28.0 mL of HNO3.

Express your answer numerically.

Part D

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 33.0 mL of NaOH.

Express your answer numerically.

Answer 1

Part C

Write down the balanced chemical equation for the reaction between NH₃ and HNO₃ as

NH₃ (aq) + HNO₃ (aq) ---------> NH₄⁺NO₃^- (aq)

NH₄⁺ is the conjugate acid of the weak base NH₃.

As per the stoichiometric equation,

1 mole NH₃ = 1 mole HNO₃ = 1 mole NH₄⁺.

Millimoles of NH₃ added = (75 mL)*(0.200 M) = 15.000 mmole.

Millimoles of HNO₃ added = (28 mL)*(0.500 M) = 14.000 mmole.

Millimoles of NH₃ neutralized by 14.000 mmole HNO₃ = millimoles of NH₄⁺ formed = 14.000 mmole.

Millimoles of NH₃ retained after the neutralization step = (15.000 – 14.000) mmole = 1.000 mmole.

Since the volume of the solution stays constant, we can express the ratio of the molarities of NH₄⁺ and NH₃ as the ratio of the number of millimoles, i.e,

[NH₄⁺]/[NH₃] = (14.000 mmole)/(1.000 mmole) = 14.000.

Use Henderson-Hasslebach’s equation for bases to find the pOH of the solution as

pOH = pK_b + log [NH₄⁺]/[NH₃]

= -log (K_b) + log (14.000)

= -log (1.8*10^-5) + 1.146

= 4.745 + 1.146 = 5.900

We know that pH + pOH = 14; therefore,

pH = 14 – pOH = 14 – 5.900 = 8.100 ≈ 8.1 (ans).

Part D

Write down the balanced chemical equation for the reaction between CH₃COOH and NaOH as

CH₃COOH (aq) + NaOH (aq) ---------> CH₃COO^-Na⁺ (aq) + H₂O (l)

CH₃COO^- is the conjugate base of the weak acid CH₃COOH.

As per the stoichiometric equation,

1 mole CH₃COOH = 1 mole NaOH = 1 mole CH₃COO^-

Millimoles of CH₃COOH added = (52.0 mL)*(0.35 M) = 18.2 mmole.

Millimoles of NaOH added = (33.0 mL)*(0.40 M) = 13.2 mmole.

Millimoles of CH₃COOH neutralized by 13.2 mmole NaOH = millimoles of CH₃COO^- formed = 13.2 mmole.

Millimoles of CH₃COOH retained after the neutralization step = (18.2 – 13.2) mmole = 5.0 mmole.

Since the volume of the solution stays constant, we can express the ratio of the molarities of CH₃COOH and CH₃COO^- as the ratio of the number of millimoles, i.e,

[CH₃COO^-]/[CH₃COOH] = (13.2 mmole)/(5.0 mmole) = 2.64

Use Henderson-Hasslebach’s equation for acids to find the pH of the solution as

pH = pK_a + log [CH₃COO^-]/[CH₃COOH]

= -log (K_a) + log (2.64)

= -log (1.8*10^-5) + 0.422

= 4.745 + 0.422 = 5.167 ≈ 5.2 (ans).