25.0 mL of 0.275 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?

Question

25.0 mL of 0.275 M  HNO₂ is titrated to its equivalence point with 1.00 M  NaOH. What is the pH at the equivalence point?

Answer 1

Mole Acid = Mole Base at equivalence point

So that M₁ V₁ = M₂ V₂ since 1 mole base is reacting with 1 mole acid based on the reaction between HNO₂ and NaOH.

HNO₂ + NaOH => NaNO₂ + H₂O

M₁ V₁ = M₂ V₂

25.0 mL x 0.275M = 1.00M x V₂ Molarity units (M) cancel each other on both side and answer will be in mL.

V₂ = V_NaOH = 6.88mL = 0.00688L

When reaction is completed there will be no HNO₂ and NaOH will be left over. Based on stoichiometry mol NaNO₂ formed will be equal to mol HNO₂.

25.0mL x 1L/1000mL = 0.0250L

M= 0.275M(mol/L)

mol HNO₂= 0.0250L x 0.275mol/L = 0.00687mol

Volume of reaction after reaction = 0.0250L + 0.00688L = 0.0319L

Molarity of HNO₂=0.00687mol/0.0319L = 0.215M

Conjugate base NO₂^- will react with water based on the reaction below.

Reaction NO₂^-_(aq) + H₂O ===> HNO₂ + OH^-

Initial 0.215M - 0 0

Change -x - +x +x

Equilibrium 0.0215 - x x x

K_b of NO₂^- = 2.2 x 10^-11 whic is not given in the question but it must be given. Instead of K_b, K_a could be given but you would have to convert it to K_b with this formula first. K_a x K_b = 1 x 10^-14

K_b = [HNO₂] [OH^-] / [NO₂^-]

2.2 x 10^-11 = x ^. x / 0.0215-x since K_b value is so little x in (0.0215-x) will be neglected.

2.2 x 10^-11 = x ^. x / 0.0215

x = [OH^-] = 6.87 x 10^-6

pOH = -log [OH^-] = -log (6.87 x 10^-6) = 5.164

pH + pOH = 14 pH = 14 - 5.164 pH = 8.836

Answer 2

At the equivalence point, moles of HNO₂ equals moles of NaOH.

HNO₂ + NaOH ==> NaNO₂ + H₂O

HNO₂ + OH^- = NO₂^- + H₂O ... net ionic equation

moles HNO₂ present = 25.0 ml x 1 L / 1000 ml x 0.275 mol / L = 0.00688 mol

moles NaOH needed = 0.00688 mol

volume NaOH needed = 0.00688 mol x 1 L / 1.00 mol = 0.00688 L (6.88 ml)

Final volume @ equivalence = 25.0 ml + 6.88 ml = 31.88 ml = 0.03188 L

Moles NO₂^- formed = 0.00688 mol (see stoichiometry of net ionic equation)

Final [NO₂^-] = 0.00688 mol / 0.03188 L = 0.216 M

NO₂^- + H₂O ==> HNO₂ + OH^-

Kb = [HNO₂][OH^-] / [NO₂^-] and Kb for NO₂^- = 1x10^-14 / 4.0x10^-4 = 2.5x10^-11

2.5x10^-11 = (x)(x) / 0.216 - x (assume x is small and ignore it in denominator)

2.5x10^-11 = x² / 0.216

x = [OH^-] = 2.32x10^-6 M

pOH = -log 2.32x10^-6 = 5.63

pH = 14 - 5.63

pH = 8.37