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JAMB Chemistry 2019 Paper

Question : 42

Total: 80

^{3}

of 1.25 molar tetraoxosulphate (vi) acid?

Solution:

Equation of reaction : 2NaOH + H $_{2}$ SO $_{4}$ → Na $_{2}$ SO $_{4}$ + 2H $_{2}$ O

Concentration of a base, CB = 0.5M

Volume of acid, V

_{A}

= 10cm

^{3}

Concentration of an acid, C

_{A}

= 1.25M

Volume of base, V

_{B}

= ?

Recall:

$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$ ... (1)

N.B: From the equation,

$\frac{n_{A}}{n_{B}} = \frac{1}{2}$

From (1)

$\frac{1.25 \times 10}{0.5 \times V_{B}} = \frac{1}{2}$

$\frac{12.5}{0.5V_{B}} = \frac{1}{2}$

25 = 0.5V $_{B}$

VB = 50.0 cm

^{3}

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