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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\(^3\)

The correct answer is 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\(^3\)


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Discussion (2)

  • Option [D] is the appropriate answer.

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  • 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\(^3\)

    Reply
    Like