Question

a solution of nitrous acid (HNO2 , Ka=4.5*10^-4) has a pH OF 1.85. How many grams of nitrous acid are there in 100 ml of the solution?

Answer 1

Here the pH and Ka of HNO₂ solution is given,

pH = 1.85

pKa = -log Ka = - log ( 4.5 x 10^-4 ) = 3.3467

We can use Henderson-Hasselbalch equation for finding out the concentration.

If an acid dissociate into ions, HA H⁺ + A^-

pH = pKa + log ( A^- / HA )

Here the dissociation of HNO₂ is, HNO₂   H⁺ + NO₂^-

log ( [A^- ] / [HA] ) = pH - pKa = 1.85 - 3.3467 = -1.4967

( [A^- ] / [HA] ) = ( [NO₂^- ] / [ HNO₂ ] ) = antilog ( -1.4967) = 0.03186

concentration of H⁺ = 10^-pH = 10^-1.85 = 0.0141 M

Here concentration of H⁺ is equal to concentration of NO₂^- as per chemical equation.

[NO₂^- ] = 0.0141 M

( [NO₂^- ] / [ HNO₂ ] ) = 0.03186

[ HNO₂ ] = 0.0141 / 0.03186 = 0.4425 M

Molarity = No. of moles of solute in 1 L of solution

= no. of moles of HNO₂ / volume in L

Volume = 100 ml = 0.1 L

No. of moles = molarity x volume = 0.4425 x 0.1 = 0.04425 moles

Molar mass of nitrous acid = 47 g/mol

Mass in gram of nitrous acid = no. of moles x molar mass = 0.04425 x 47 = 2.08 g

Hence 2.08 g of nitrous acid is present in 100 ml of the solution.

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