Step-by-step 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%