The volume of 0.5 M `H_(3)PO_(4)` that completely dissolved 3.1 g of copper carbonate is (molecular mass of copper carbonate`=124gmol^(-1))`

Options

A. 55.5 mL
B. 45.5 mL
C. 35.5 mL
D. 33.3 mL

Explanation

To find the volume of 0.5 M H₃PO₄ that completely dissolves 3.1 g of copper carbonate (CuCO₃), we can follow these steps:

### Step 1: Calculate the number of moles of CuCO₃
We start by calculating the number of moles of copper carbonate using its mass and molecular weight.

\[ \text{Moles of CuCO}_3 = \frac{\text{mass}}{\text{molecular mass}} = \frac{3.1 \, \text{g}}{124 \, \text{g/mol}} = 0.025 \, \text{mol} \]

### Step 2: Determine the equivalent of CuCO₃
Next, we need to find the number of equivalents of CuCO₃. The equivalent weight is calculated as:

\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n \text{ factor}} \]

For CuCO₃, the n factor is 2 (as it can produce two moles of ions). Therefore:

\[ \text{Equivalents of CuCO}_3 = \frac{3.1 \, \text{g}}{124 \, \text{g/mol} / 2} = \frac{3.1}{62} = 0.050 \, \text{equivalents} \]

### Step 3: Determine the equivalents of H₃PO₄ needed
H₃PO₄ has an n factor of 3 because it can donate three protons (H⁺ ions). Therefore, the number of equivalents of H₃PO₄ needed will be equal to the equivalents of CuCO₃:

\[ \text{Equivalents of H}_3PO_4 = 0.050 \, \text{equivalents} \]

### Step 4: Calculate the volume of H₃PO₄ solution
Using the molarity (M) and the relationship between moles, volume, and equivalents, we can find the volume of H₃PO₄ needed:

\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (L)} \]

Rearranging gives us:

\[ \text{Volume (L)} = \frac{\text{Equivalents}}{\text{Molarity}} = \frac{0.050}{0.5} = 0.1 \, \text{L} \]

To convert this to milliliters:

\[ \text{Volume (mL)} = 0.1 \, \text{L} \times 1000 \, \text{mL/L} = 100 \, \text{mL} \]

### Final Calculation
However, we need to consider the n factor of H₃PO₄. Since it can donate 3 protons, we need to divide the volume by the n factor:

\[ \text{Volume of H}_3PO_4 = \frac{100 \, \text{mL}}{3} \approx 33.33 \, \text{mL} \]

### Conclusion
Thus, the volume of 0.5 M H₃PO₄ that completely dissolves 3.1 g of copper carbonate is approximately **33.3 mL**.

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