Question

What is the equivalent weight of \( Na_2CO_3 \) in the reaction, represented by \( Na_2CO_3 + HCl \rightarrow NaHCO_3 + NaCl \)?

• A. \(\text{53}\)
• B. \(\text{5.3}\)
• C. \(\text{106}\)
• D. \(\text{10.6}\)

Hint:

Equivalent weight is found by: \[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{Number of replaceable } H^+ \text{ ions accepted or donated}} \]

Answer 1

The Correct Option is A

Step 1: Understanding equivalent weight. 
- Equivalent weight is given by: \[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{Number of replaceable } H^+ \text{ ions accepted or donated}} \]

\[ \text{Molar mass of } Na_2CO_3 = (23 \times 2) + 12 + (16 \times 3) = 106 \, \text{g/mol} \]

\[ \frac{106}{2} = 53 \, \text{g/mol} \]

- Since \( Na_2CO_3 \) reacts with one mole of HCl, it donates one \( CO_3^{2-} \) per mole. - Equivalent weight: \[ \frac{106}{2} = 53 \] 
Step 2: Selecting the correct option. Since the equivalent weight of \( Na_2CO_3 \) is 53 g/mol, the correct answer is (A) 53.