Determine the ph at the equivalence point for the titration of 40.0 ml of 0.1 m hno2 by 0.200 m koh? (the pka of nitrous acid is 3.36.)

pH = 8.092

Explanation:

The equivalence point, or stoichiometric point, of a chemical reaction is the point at which chemically equivalent quantities of reactants have been mixed. In other words, the moles of acid are equivalent to the moles of base, according to the equation.

First thing'ss first, we have to write out the chemical equaion;

HNO₂(aq) + KOH(aq) → KNO₂(aq) + H₂O(l)

From the question, we were given;

Volume of HNO2 = 40ml

Molarity = 0.1M

Number of moles = Molarity * Volume

No. of moles of HNO₂ = (0.1 mol/L) × (40.0/1000 L) = 0.004 mol

From the equation;

Mole ratio HNO₂ : KOH = 1 : 1

Hence;

No. of moles of KOH = 0.004 mol

But, Volume = Number of moles / Molariy

Volume of KOH = (0.004 mol) / (0.200 mol/L) = 0.02 L

From KNO₂ ⇄ K⁺ + NO₂⁻

1 mole of KNO₂ contains 1 mole of NO₂⁻ ions.

No. of moles of NO₂⁻ ions = 0.004 mol

Volume of the final solution = (40.0/1000) + (0.02 L) = 0.06 L

Molarity of NO₂⁻ ions in the final solution = (0.004 mol) / (0.06 L) = 0.06667 M

Consider the dissociation of NO₂⁻ :

NO₂⁻(aq) + H₂O(l) ⇌ HNO₂(aq) + OH⁻(aq) Kb

Kb = Concentration of products / Concentration of reactants

Kb = [HNO₂][OH⁻] / [NO₂⁻] =

Kb = ([H⁺][OH⁻]) / [NO₂⁻][H⁺]/[HNO₂]

Kb = Kw/Ka

The pKa of nitrous acid is 3.36.

The pKa of water is 14.

Kb = (1 × 10⁻¹⁴)/(10⁻³·³⁶)

Before the dissociation :

[NO₂⁻]ₒ = 0.06667 M

[HNO₂]ₒ = [OH⁻]ₒ = 0 M

At equilibrium :

[NO₂⁻] = (0.06667 - y) M ≈ 0.06667 M, assuming that 0.06667 ≫ y

[HNO₂] = [OH⁻] = y M

[HNO₂][OH⁻]/[NO₂⁻] = Kb

y² / 0.06667 = (1 × 10⁻¹⁴)/(10⁻³·³⁶)

y = 1.236 × 10⁻⁶

[OH⁻] = 1.236 × 10⁻⁶ M

pH = 14 - pOH = 14 - (-log[OH⁻] = 14 + log(1.236 × 10⁻⁶) = 8.092

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