Acid-base titrations Questions

The Acid-base titrations work well together. In the chemical process, the acid loses a proton and the base gains a proton. The acid-base interaction produces salt and water.

Titration, defined as the determination of an unknown concentration of an acid or base using an acid or base of known concentration, is the method by which these concentrations are determined. In acid-base titrations, pH-sensitive indicators are employed.

Acid-base titrations Chemistry Questions with Solutions

Q1. What is the common-ion effect?

Ans – The ionization equilibrium of an acid or base is upset by the addition of another acid or base that shares one ion with the initial acid or base. In consequence, the original compound’s ionization reduces. This effect is known as the common ion effect.

Q2. How do the buffer solutions resist the change in pH?

Ans – A buffer is a solution containing either a high concentration of weak acid and its conjugate base or a high concentration of weak base and its conjugate acid.

Both of these buffers operate on the same principle: if acid is introduced to a weak acid-conjugate base buffer solution, the conjugate base neutralizes it. The weak acid neutralizes the buffer if a base is supplied, and vice versa.

If the conjugate base/weak acid ratio does not change significantly towards the end, the pH value of the solution does not change significantly.

Q3. Differences between strong and weak acids and give 2 examples of each.

Ans – The differences between the strong and weak acids are given below:

S. No.	Strong Acids	Weak Acids
1.	Dissociate 100% in aqueous solution.	Do not dissociate 100% in an aqueous solution.
2.	pH of strong acids lie in the range 1-2.5	pH of weak acid lie in the range 3-5
3.	Ka value is higher.	Ka value is lower.
4.	H+ ions dissociate completely.	H+ ions do not dissociate completely.
5.	Examples are: HCl and H2SO4	Examples are: CH3COOH and HCOOH

Q4. In the titration of strong acid vs strong base, what trend of pH is observed?

Ans – Near the equivalency point, the pH of the solution increases rapidly.

Q5. Why are the solutions of FeCl₃ and Na₂CO₃ not neutral?

Ans – FeCl3 is a salt of HCl, which is a strong acid, and ferric hydroxide, which is a weak base. Therefore, in solution, FeCl3 undergoes hydrolysis and generates HCl, a powerful acid that completely dissociates into the solution, rendering it acidic. This is the reason why FeCl3 solution is acidic.

FeCl3 + 3H2O → Fe(OH)3 + 3HCl

Similarly, Na2CO3, which is a salt of the strong base NaOH and the weak acid H2CO3, is hydrolyzed to create NaOH, which raises the solution’s pH.

Na2CO3 +H2O -> NaOH + H2CO3

Therefore, Na2CO3 solution is basic in nature.

Q6. What will be the pH of the solution when 25 mL of 0.1 M NaOH is added to 40 mL of 0.1 M HCl solution?

Ans – Given: volume of NaOH (V₁) = 25 mL, molarity of NaOH solution (M₁) = 0.1 M

Number of moles of NaOH = n₁

Volume of HCl (V₂) = 40 mL, molarity of HCl solution (M₂) = 0.1 M

Number of moles of HCl = n₂

Now, Number of moles of NaOH, n₁ = M₁ x V₁ = 0.025 L x 0.1 mol/L = 0.0025 mol

Number of moles of HCl, n₂ = M₂ x V₂ = 0.040 L x 0.1 mol/L = 0.004 mol

HCl and NaOH react in a 1:1 ratio. Hence, HCl moles are in excess and will turn the solution acidic.

Now, the remaining moles of HCl in the solution, n = n₂ – n₁ = 0.004 – 0.0025 = 0.0015 mol

Molarity = No. of moles / Volume of solution (L)

Total Volume of the solution = 0.025 L + 0.04 L = 0.065 L

Hence, molarity of the remaining solution= 0.0015 / 0.065 mol L^-1 = 0.023 M

Now, pH = -log [H⁺] = -log[0.023] = 1.638

Hence, the pH of the solution after the reaction is 1.64.

Q7. What will be the pH at the equivalence point of the reaction when 50 mL of 0.257 M HBr is titrated with 0.450 M KOH solution?

Ans – Both HBr and KOH are extremely strong acids and bases. At the equivalence point, the strong acid and strong base neutralize fully, and the solution becomes neutral. Therefore, pH must be 7.

Q8. What is the molarity of a solution of sodium hydroxide, NaOH, if 41.4 mL of this solution is required to react with 37.5 mL of 0.0342 M nitric acid solution according to the following reaction?

HNO3 + NaOH→NaNO3 + H2O(ℓ)

Ans – From the volume and concentration of the nitric acid solution, we can solve for the number of moles of nitric acid that reacted with sodium hydroxide.

We will then use the mole ratio of nitric acid and sodium hydroxide which is 1:1 to convert it to moles of sodium hydroxide.

The concentration of the sodium hydroxide solution is 0.03098 M.

Q9. If 83 mL of 0.45 M NaOH solution neutralizes a 235 mL HCl solution. Calculate the molarity of the HCl solution.

Ans – The balanced chemical reaction between HCl and NaOH is as follows:

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)

From the molarity equation: M_AV_A / n_A (acid) = M_BV_B / n_B (base)

Where: n_A and n_B are the number of moles from the balanced chemical equation.

M_A and V_A are the molarity and volume of the acid

M_B and V_B are the molarity and volume of the base.

Hence, M_A = M_BV_Bn_A / n_BV_A = 0.45 M x 83 mL x 1 / 235 mL x 1 = 0.16 M

The molarity of the HCl solution is 0.16 M.