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## Q. 4.4

An aqueous solution called Kalkwasser $[0.0225 M Ca(OH)_{2}]$ may be added to saltwater aquaria to reduce acidity and add $Ca^{2+}$ ions. How many grams of $Ca(OH)_{2}$ do you need to make 500.0 mL of Kalkwasser?

## Verified Solution

Collect, Organize, and Analyze We know the volume and molarity of the solution we must prepare. We also know the identity of the solute, which enables us to calculate its molar mass. We use Equation 4.2

$m_{solute}=(V\times M)\times \mathscr{M}$              (4.2)

to calculate the mass of $Ca(OH)_{2}$ needed. We also must convert the volume from milliliters to liters to obtain the mass, in grams, of solute required.

The molar mass of $Ca(OH)_{2}$ is

$40.08 g/mol + 2(16.00 g/mol) + 2(1.008 g/mol) = 74.10 g/mol$

Solve Equation 4.2 can be used to calculate the mass of Ca(OH)2 needed:

$m=V \times M\times \mathscr{M}$

$=500.0 \sout{mL} \times \frac{1 \sout{L}}{10^{3} \sout{mL}} \times \frac{0.0225 \sout{mol}}{1 \sout{L}} \times \frac{74.10 g}{1 \sout{mol}} =0.834 g$

Figure 4.6 shows the technique for making this solution. A key feature in the preparation is that we suspend 0.834 g of solute in 200–300 mL of water and then add additional water until the final solution has the volume specified.

Think About It To solve a problem like this, all we need to remember is the definition of molarity: moles of solute per liter of solution. The solution we want has a concentration of 0.0225 M, or 0.0225 mol/L. Because we need only 0.5000 L of this solution, we do not need 0.0225 mol of $Ca(OH)_{2}$, but only half that amount (0.0112 mol). Finally, we calculate the number of grams of calcium hydroxide in 0.0112 mol of calcium hydroxide (0.834 g). The amount of $Ca(OH)_{2}$ needed is less than a teaspoon.