To determine how many millilitres of 0.5 M $ \text{H}_2\text{SO}_4 $ are needed to dissolve 0.5 g of copper(II) carbonate $ (\text{CuCO}_3) $, we first need to write the chemical reaction between the copper(II) carbonate and sulfuric acid ( $ \text{H}_2\text{SO}_4 $ ):

$ \text{CuCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CuSO}_4 + \text{H}_2\text{O} + \text{CO}_2 $

From this reaction, it's observable that 1 mole of copper(II) carbonate reacts with 1 mole of sulfuric acid to produce 1 mole of copper(II) sulfate, water, and carbon dioxide.

Let's calculate the number of moles of copper(II) carbonate in 0.5 g:

The molar mass of $ \text{CuCO}_3 $ is calculated as:

$ \text{Molar mass of } \text{CuCO}_3 = (\text{Molar mass of Cu}) + (\text{Molar mass of C}) + 3 \times (\text{Molar mass of O}) $

$ \text{Molar mass of } \text{CuCO}_3 = (63.55 \text{ g/mol}) + (12.01 \text{ g/mol}) + 3 \times (16.00\text{ g/mol}) $

$ \text{Molar mass of } \text{CuCO}_3 = 63.55 + 12.01 + 48.00 $

$ \text{Molar mass of } \text{CuCO}_3 = 123.56 \text{ g/mol} $

Now, we'll find out the number of moles in 0.5 g of $ \text{CuCO}_3 $:

$ \text{Number of moles of } \text{CuCO}_3 = \frac{\text{mass of } \text{CuCO}_3}{\text{Molar mass of } \text{CuCO}_3} $

$ \text{Number of moles of } \text{CuCO}_3 = \frac{0.5 \text{ g}}{123.56 \text{ g/mol}} $

$ \text{Number of moles of } \text{CuCO}_3 \approx 0.004046 \text{ mol} $

Since the reaction is a 1:1 molar ratio, the moles of sulfuric acid needed would also be $ 0.004046 \text{ mol} $.

The concentration (C) of a solution is given by the formula:

$ C = \frac{n}{V} $

where:

• $ C $ is the concentration of the solution in moles per liter (M).


• $ n $ is the number of moles of solute.


• $ V $ is the volume of solution in liters.

Rearranging the formula to solve for volume $ V $:

$ V = \frac{n}{C} $

The given concentration of $ \text{H}_2\text{SO}_4 $ is 0.5 M, which means 0.5 moles of $ \text{H}_2\text{SO}_4 $ per liter. Now, let's calculate the volume required for the $ 0.004046 \text{ mol} $ of $ \text{H}_2\text{SO}_4 $:

$ V = \frac{0.004046 \text{ mol}}{0.5 \text{ M}} = \frac{0.004046 \text{ mol}}{0.5 \text{ mol/L}} $

$ V = 0.008092 \text{ L} $

To convert liters to millilitres:

$ V(\text{mL}) = 0.008092 \text{ L} \times 1000 \frac{\text{mL}}{\text{L}} $

$ V(\text{mL}) = 8.092 \text{ mL} $

Therefore, you will need approximately 8.092 mL of 0.5 M $ \text{H}_2\text{SO}_4 $ to completely dissolve 0.5 g of copper(II) carbonate under the assumption that the reaction goes to completion with no side reactions.