1.725 g of a metal carbonate is mixed with 300 mL of `N/10` HCI. 10 mL of `N/2` sodium hydroxide were required to neutralise excess of the acid. Calculate the equivalent mass of the metal carbonate.

Answer

To solve the problem step-by-step, we will follow the outlined procedure to calculate the equivalent mass of the metal carbonate.

### Step 1: Understand the given data
We have:
- Mass of metal carbonate = 1.725 g
- Volume of HCl = 300 mL
- Normality of HCl = N/10 (which is 0.1 N)
- Volume of NaOH used to neutralize excess HCl = 10 mL
- Normality of NaOH = N/2 (which is 0.5 N)

### Step 2: Calculate the number of equivalents of NaOH used
The number of equivalents of NaOH can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (in L)} \]
For NaOH:
\[ \text{Number of equivalents of NaOH} = 0.5 \, \text{N} \times 0.010 \, \text{L} = 0.005 \, \text{equivalents} \]

### Step 3: Determine the equivalents of HCl neutralized by NaOH
Since NaOH neutralizes HCl, the number of equivalents of HCl neutralized by NaOH is equal to the number of equivalents of NaOH:
\[ \text{Number of equivalents of HCl neutralized} = 0.005 \, \text{equivalents} \]

### Step 4: Calculate the total number of equivalents of HCl initially present
The total number of equivalents of HCl in 300 mL of N/10 HCl:
\[ \text{Total equivalents of HCl} = \text{Normality} \times \text{Volume (in L)} = 0.1 \, \text{N} \times 0.300 \, \text{L} = 0.03 \, \text{equivalents} \]

### Step 5: Calculate the number of equivalents of HCl that reacted with the metal carbonate
The equivalents of HCl that reacted with the metal carbonate can be calculated as:
\[ \text{Equivalents of HCl reacted} = \text{Total equivalents of HCl} - \text{Equivalents of HCl neutralized by NaOH} \]
\[ \text{Equivalents of HCl reacted} = 0.03 - 0.005 = 0.025 \, \text{equivalents} \]

### Step 6: Calculate the equivalent mass of the metal carbonate
The equivalent mass of the metal carbonate can be calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Mass of metal carbonate}}{\text{Number of equivalents of HCl reacted}} \]
\[ \text{Equivalent mass} = \frac{1.725 \, \text{g}}{0.025 \, \text{equivalents}} = 69 \, \text{g/equivalent} \]

### Final Answer
The equivalent mass of the metal carbonate is **69 g/equivalent**.

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