Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 4

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?

Step-by-Step

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.

Figure 4.6