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?
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Correct Answer: Option D
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\(^3\)
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\(^3\)