20cm3 of hydrochloric acid neutralized 25cm3 0.1moldm-3 sodium trioxocarbonate(IV) solution, find (i) the concentration of hydrochloric acid in moldm-3. (ii) the concentration of hydrochloric acid in gdm-3. (H=1.00, Cl=35.5)The equation for the reaction: 2HCl + Na2 CO3 → 2NaCl + H2O

1. 0.25 moldm¯³.

2. 9.125 gdm¯³.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2HCl + Na2CO3 → 2NaCl + H2O + CO2

From the balanced equation,

The mole ratio of the acid, HCl (nA) = 2

The mole ratio of the base, Na2CO3 (nB) = 1

1. Determination of the concentration of the acid, HCl in moldm¯³.

Volume of acid, HCl (Va) = 20 cm³

Volume of base, Na2CO3 (Va) = 25 cm³

Concentration of base, Na2CO3 (Cb) = 0.1 moldm¯³

The mole ratio of the acid, HCl (nA) = 2

The mole ratio of the base, Na2CO3 (nB) = 1

Concentration of acid, HCl (Ca) =?

Cava /CbVb = nA/nB

Ca × 20 / 0.1 × 25 = 2/1

Ca × 20 / 2.5 = 2

Cross multiply

Ca × 20 = 2.5 × 2

Ca × 20 = 5

Divide both side by 20

Ca = 5/20

Ca = 0.25 moldm¯³

Thus, the concentration of the acid, HCl in moldm¯³ is 0.25 moldm¯³

2. Determination of the concentration of the acid, HCl in gdm¯³

Concentration of the acid, HCl = 0.25 moldm¯³

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Concentration of the acid, HCl in gdm¯³ =?

Conc. in moldm¯³ = Conc. in gdm¯³ /Molar mass

0.25 = Conc. in gdm¯³ /36.5

Cross multiply

Conc. in gdm¯³ = 0.25 × 36.5

Conc. in gdm¯³ = 9.125 gdm¯³.

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