Answer to Question #189474 in General Chemistry for Jowheecelle

Question

A volume of 25.0 mL of 1.9752M strong monoprotic acid, HA, is to be titrated with 0.4115M strong base, B(OH)2. Answer the questions that follow and Write answer in THREE SIGNIFICANT FIGURES with complete solution.

A. Calculate the volume of the base titrant (in milliliters) required to reach the equivalence point.

B. Calculate the pH of the resulting solution when the original solution has been titrated with 50.0 mL of the base titrant

Calculate the pH of the resulting solution when the original solution has been titrated with 50.0 mL of the base titrant.

C. Calculate the pH of the resulting solution when the original solution has been titrated with 55.0 mL of the base titrant. 

D. Calculate the pH of the resulting solution when the original solution has been titrated with 60.0 mL of the base titrant.

E. Calculate the pH of the resulting solution when the original solution has been titrated with 65.0 mL of the base titrant.

Answer 1

Step 1

Volume titrant can be calculated using law of chemical equivalence:

n₂ × M₂ × V₂ = n₁ × M₁ × V₁ 

Step 2

Answer:

A) given molarity of HA = 1.9752 M

n-factor for HA = 1 (since it is monoprotic acid)

Volume of HA = 25.0 mL

Molarity of base, B(OH)₂ = 0.4115 M

n-factor for B(OH)₂ = 2 (since it's a dibasic base)

Volume of B(OH)₂ = V

Applying equation of chemical equivalence:

(n × M × V)_B(OH)2 = (n × M × V)_HA 

=> 2 × 0.4115 M × V = 1 × 1.9752 M × 25.00 mL

=> V = 60.00 mL

So the volume of base titrant, B(OH)₂required at equivalence point is 60.0 mL

B) Given 50.0 mL of the base titrant added

=> Total volume of solution = volume of acid + volume of base = 25.0 mL + 50.0 mL = 75.0 mL = 0.075 L (since 1 mL = 0.001 L)

Now the mmol of base added = n-factor × Molarity × volume = 2 × 0.4115 M × 50.0 mL = 41.15 mmol

mmol of acid present = 1 × 1.9752 M × 25.0 mL = 49.38 mmol

=> Excess mmol of acid remaining after neutralization = mmol of acid initially - mmol of base added = 49.38 mmol - 41.15 mmol = 8.23 mmol

=> Molarity of strong acid, HA = 8.23 mmol / 75.0 mL = 0.1097 M

Since pH = -log[H⁺]

=> pH = -log(0.1097 M) = 0.959 

So the pH of the resulting solution when the original solution has been titrated with 50.0 mL of the base titrant is 0.959 

C) Given 55.0 mL of the base titrant added

=> Total volume of solution = volume of acid + volume of base = 25.0 mL + 55.0 mL = 80.0 mL

Now the mmol of base added = n-factor × Molarity × volume = 2 × 0.4115 M × 55.0 mL = 45.265 mmol

mmol of acid present = 1 × 1.9752 M × 25.0 mL = 49.38 mmol

=> Excess mmol of base present after neutralization = mmol of acid initially - mmol of base added = 49.38 mmol - 45.265 mmol = 4.115 mmol

=> Molarity of strong acid, HA = 4.115 mmol / 80.0 mL = 0.0514 M

Since pH = -log[H⁺]

=> pH = -log(0.0514 M) = 1.29

So the pH of the resulting solution when the original solution has been titrated with 55.0 mL of the base titrant is 1.29

D) at equivalence point 60.0 mL there is salt of strong acid and strong base, whose pH is equal to the pH of water, i.e., 7

 

pH above neutralization points i.e. above 60.0 mL of Base can be calculated by excess mmol of base added then convert them to Molarity don't forget 1 mmol of B(OH)₂ gives 2 mmol of OH^-, then calculate pOH, and pH = 14 - pOH