Question

What volume (in mL) of a 0.150 M HNO₃ solution will completely react with 35.7 mL of a 0.108 M Na₂CO₃ solution according to this balanced chemical equation? $Na_2CO_3(aq) + 2HNO_3(aq) \rightarrow 2NaNO_3(aq) + CO_2(g) + H_2O(l)$

Answer 1

Answer: 51.408ml

Explanation: The balanced chemical equation

(Tex)Na2CO3 (aq) + 2HNO3 (aq) 2NaNO3 (aq) + CO2 (g) + H2O (l) (Tex)

Na2CO3 ionizes as

Na2CO3 2Na+ + CO32-

Water simply ionizes as

H2O H+ + OH-

The ion CO32- reacts with water

(Tex) CO32- + H+ + OH- H2CO3 + OH-(Tex)

The presence of excess OH- makes a solution of Na2CO3 basic or alkaline, so its reaction with HNO3 can be termed an acid-base reaction.

We use the equation for the calculation:

CaVa/CbVb = na / nb

Where Ca = molar concentration or molarity of the acid, HNO3 = 0.150M,

Cb = molar concentration of the base, Na2CO3 = 0.108M,

Va = volume of the acid = ?

Vb = volume of the base = 35.7ml

na = mole ratio of the acid = 2 ( the stoichiometric coefficient of HNO3 from the balanced equation )

nb= mole ratio of the base = 1 ( the stoichiometric coefficient of Na2CO3 from the balanced equation )

Va = CbVbna / Canb = 0.108 x 35.7 x 2 / 0.150 x 1 = 7.7112 / 0.150 = 51.408ml