Concept explainers
(a)
Interpretation:
The concentration of potassium ions in the solution that is prepared has to be calculated.
Concept Introduction:
Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is mol⋅L−1. The equation for molarity is given as follows;
Molarity (M)=Amount of solute in molesVolume of solution in liters
From the above equation, the mass of solute can be calculated if the concentration and volume is known.
(a)
Answer to Problem G.21E
The concentration of potassium ions is 4.58×10−2 M.
Explanation of Solution
The solution is said to be prepared by dissolving 0.500 g of KCl, 0.500 g of K2S, and 0.500 g of K3PO4 in 500.0 mL of water. The number of moles of potassium ions that is contributed by each component is calculated as follows;
Number of moles of potassium from 0.500 g of KCl:
Molar mass of KCl is 74.5 g⋅mol−1. Number of moles of KCl is calculated as follows;
n(KCl)=0.500 g74.5 g⋅mol−1=0.0067 mol
One mole of KCl gives one mole of potassium ions. Therefore, 0.0067 mol of KCl gives 0.0067 mol of potassium ions.
Number of moles of potassium from 0.500 g of K2S:
Molar mass of K2S is 110.3 g⋅mol−1. Number of moles of K2S is calculated as follows;
n(K2S)=0.500 g110.3 g⋅mol−1=0.0045 mol
One mole of K2S gives two moles of potassium ions. Therefore, 0.0045 mol of K2S gives 0.0090 mol of potassium ions.
Number of moles of potassium from 0.500 g of K3PO4:
Molar mass of K3PO4 is 212.27 g⋅mol−1. Number of moles of K3PO4 is calculated as follows;
n(K3PO4)=0.500 g212.27 g⋅mol−1=0.0024 mol
One mole of K3PO4 gives three moles of potassium ions. Therefore, 0.0024 mol of K3PO4 gives 0.0072 mol of potassium ions.
Thus, the total potassium ions in the solution is calculated by summing up all the moles of potassium ions. This is done as follows;
Total No. of K+ ions=0.0067 mol+0.0090 mol+0.0072 mol=0.0229 mol
The volume of the solution is given as 0.500 L. Therefore, the concentration of potassium ions is calculated as shown below;
Concentration of K+ ions=0.0229 mol0.500 L=0.0458 M=4.58×10−2 M
Thus, the concentration of potassium ions in 0.500 L of solution is 4.58×10−2 M.
(b)
Interpretation:
The concentration of sulfide ions in the solution that is prepared has to be calculated.
Concept Introduction:
Refer part (a).
(b)
Answer to Problem G.21E
The concentration of sulfide ions is 9×10−3 M.
Explanation of Solution
The solution is said to be prepared by dissolving 0.500 g of KCl, 0.500 g of K2S, and 0.500 g of K3PO4 in 500.0 mL of water. The number of moles of sulfide ions that is contributed by each component is calculated as follows;
Among the three components, only sulfide ion is contributed by K2S. Therefore, the number of moles of sulfide from 0.500 g of K2S:
Molar mass of K2S is 110.3 g⋅mol−1. Number of moles of K2S is calculated as follows;
n(K2S)=0.500 g110.3 g⋅mol−1=0.0045 mol
One mole of K2S gives one mole of sulfide ions. Therefore, 0.0045 mol of K2S gives 0.0045 mol of sulfide ions.
The volume of the solution is given as 0.500 L. Therefore, the concentration of sulfide ions is calculated as shown below;
Concentration of S2- ions=0.0045 mol0.500 L=0.009 M=9×10−3 M
Thus, the concentration of sulfide ions in 0.500 L of solution is 9×10−3 M.
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Chapter F Solutions
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 1TERM
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