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?

  • A 25cm\(^3\)
  • B 10cm\(^3\)
  • C 20cm\(^3\)
  • D 50cm\(^3\)
jamb 2019chemistry

The correct answer is D. 50cm\(^3\)

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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