Question

The percentage by weight of nitric acid, HNO₃, in a sample of concentrated nitric acid is to be determined.

1. Initially a NaOH solution was standardized by titration with a sample of potassium hydrogen phthalate, KHC₈H₄O₄, a monoprotic acid often used as a primary standard. A sample of pure KHC₈H₄O₄ weighing 1.518 grams was dissolved in water and titrated with the NaOH solution. To reach the equivalence point, 26.90 millilitres of base was required. Calculate the molarity of the NaOH solution. (Molecular weight: KHC₈H₄O₄ = 204.2)

b. A 10.00 millilitre sample of the concentrated nitric acid was diluted with water to a total volume of 500.0 millilitres. Then 25.00 millilitres of the diluted acid solution was titrated with the standardized NaOH solution prepared in part (a). The equivalence point was reached after 28.35 millilitres of the base had been added. Calculate the molarity of the concentrated nitric acid.

Answer 1

a.

Balanced equation is

KHC₈H₄O₄ (aq) + NaOH (aq) KNaC₈H₄O₄  (aq) + H₂O (l)

So, mole ratio of potassium hydrogen phthalate (KHP) and NaOH

= 1:1.

Now, moles of KHP

= (mass/molar mass)

= {1.518 g/204.2 (g/mol)}

= 0.0074 .

Moles of NaOH = moles of KHP = 0.0074.

Now, volume of NaOH solution

= 26.90 mL

= 26.90 mL ×

= 0.0269 L.

Therefore molarity of NaOH solution

= ( Moles of NaOH/ Volume of NaOH)

= (0.0074/0.0269)

= 0.275 M.

b.

Balanced equation is

HNO₃ + NaOH NaNO₃ + H₂O

Hence,

V_acid × M_acid = V_base × M_base

Or, 25.00 × M_acid = 28.35 × 0.275

Or, 25.00 M_acid = 7.798

Or, M_acid = (7.798/25.00)

Or, M_acid = 0.312 M.

So, molarity of dilute nitric acid is 0.312 M.

Now, let molarity of concentrated HNO₃ = M₁

initial volume (V₁) = 10.00 mL

Final volume after dilution (V₂) = 500.0 mL

Final concentration of dilute HNO₃ = 0.312 M

For dilution.

molesof HNO₃ = M₁V₁=M₂V₂

Or, M₁ × 10.00 = 0.312 × 500.0

Or, M₁ × 10.00 = 156

Or, M₁ = (156/10.00)

Or, M₁ = 15.6 M.

Therefore , molarity of concentrated nitric acid is 15.6 M.