5g of NaOH was dissolved in 1000cm3. 25cm3 of this solution neutralized 28.3cm3 of solution containing 7.2gdm-3 of impure H2SO4. Calculate I. The molarity of H2SO4 II. The concentration of the pure acid on gdm-3 III. % impurity of the acid

Explanation:

The balanced equation for the reaction between NaOH and H2SO4 is given as;

2NaOH + H2SO4 → Na2SO4 + 2H2O  

2 mol of NaOH reacts with 1 mol of H2SO4

5g of NaOH was dissolved in 1000cm3.

Mass = 5 g

Molar mass = 40 g/mol

Volume = 1000 cm3 = 1 L

Number of moles = Mass / Molar mass

Number of moles = 5 / 40 = 0.125 mol

Molarity = Number of moles / Volume

Molarity = 0.125 / 1 = 0.125 M

25cm3 of this solution neutralized 28.3cm3 of solution containing 7.2gdm-3 of impure H2SO4.

CaVa / CbVb = Na / Nb ; where a = acid and b = base

Va = 28.3cm3

Vb = 25cm3

Ca = ?

Cb = 0.125 M

Na = 1

Nb = 2

I. The molarity of H2SO4

Solving for Ca;

Ca = CbVb * Na / (Va * Nb)

Cb = (0.125 * 25 * 1 ) / ( 28.3 * 2)

Cb = 0.0552 M

II. The concentration of the pure acid on gdm-3

Molarity = Mass conc / Molar mass

Mass Conc = Molarity * Molar mass

Mass Conc = 0.0552 * 98.079

Mass Conc = 5.41 g/dm3

III. % impurity of the acid

Percentage Impurity = Mass of pure / Mass of Impure   *  100

Percentage Impurity = 5..41 / 7.2 * 100

Percentage Impurity = 75.14%

answer

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