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?

25cm $^3$

10cm $^3$

20cm $^3$

50cm $^3$

Correct answer is D

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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