Question

Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. Suppose that a solution is 0.054 M in calcium chloride and 0.093 M in magnesium nitrate. What mass of sodium phosphate would have to be added to 1.4 L of this solution to completely eliminate the hard water ions? Assume complete reaction. Enter a numerical answer only, in terms of grams to two significant figures.

Answer 1

Answer:

$22.4269$ grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

Explanation:

We will first write the balanced equation for this scenario

3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2

3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2

The ratio here for both calcium chloride and magnesium nitrate is $3:2$

The number of moles of each compound is equal to

$0.054 * 1.4 = 0.0756\\0.093* 1.4 = 0.1302$

Using the mole ratio of 3:2, convert each to moles of sodium phosphate.

$0.0756$ mole of CaCl2 is equal to $0.05$ Na3PO4

$0.1302$ mole of CaCl2 is equal to $0.0868$ Na3PO4

Converting moles of sodium phosphate to grams of sodium phosphate we get

$(0.05 +0.0868) * 163.94$ g/mol

$22.4269$ grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions