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# In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, wh...

### Question

In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm$$^3$$ of 1.25 molar tetraoxosulphate (vi) acid?

### Options

A) 25cm$$^3$$

B) 10cm$$^3$$

C) 20cm$$^3$$

D) 50cm$$^3$$

### Explanation:

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$$^3$$

## Discussion (2)

• Option [D] is the appropriate answer.

• 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$$^3$$