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\(^3\)
More Past Questions:
Discussion (2)
Other Subjects
- English Language
- Biology
- Government
- Mathematics
- Physics
- Economics
- Christian Religious Knowledge
- Commerce
- Geography
- Literature In English
- Accounts
- Agricultural Science
- General Paper
- Islamic Religious Knowledge
- History
- Further Mathematics
- Current Affairs
- Civic Education
- Computer Studies
- Yoruba
- Hausa
- Igbo
- French
- Home Economics
- Sweet Sixteen
- Fine Arts
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\(^3\)