Practice Questions Section 2.5
The Concentration of Ions in Solution

1.	Write balanced reaction equation that show which ions are produced when the following substances are dissolved in water.

a.	lithium hydroxide		
b.	potassium phosphate		
c.	strontium chloride 		
d.	chromium(III) sulfate 		

2.	Iron(III) nitrate has a solubility of 0.15 M. Find concentration of the ions in solution.

3.	Calculate ion concentrations in a 2.00 L solution containing 17.1 g aluminum sulfate, Al2(SO4)3

Practice Questions Section 2.5
The Concentration of Ions in Solution	Answers

1.	Write balanced reaction equation that show which ions are produced when the following substances are dissolved in water.

a.	lithium hydroxide	LiOH(s) → Li+(aq) + OH-(aq)	
b.	potassium phosphate	K3PO4 (s) → 3 K+(aq) + PO43-(aq)	
c.	strontium chloride 	SrCl2(s) → Sr2+(aq) + 2 Cl-(aq)	
d.	chromium(III) sulfate 	Cr2(SO4)3 (s) → 2 Cr3+(aq) + 3 SO42-(aq)	

2.	Iron(III) nitrate has a solubility of 0.15 M. Find concentration of the ions in solution.

	Solution:

		Begin by writing a balanced dissociation equation:

				Fe(NO3)3 → Fe3+(aq) + 3 NO3-(aq)

		The concentration of the ions can be determined from the balancing coefficients from the equation:

				[Fe3+] = 1× [Fe(NO3)3] = 1 × 0.15 = 0.15 M

				[NO3-] = 3× [Fe(NO3)3] = 3 × 0.15 = 0.45 M

3.	Calculate ion concentrations in a 2.00 L solution containing 17.1 g aluminum sulfate, Al2(SO4)3

	Solution:

		Before calculating the concentration of the ions, we must first calculate the concentration of the aluminum sulfate solution.
2 Al = 2 × 27.0
3 S = 3× 32.0.0
12 O = 12 × 16.0 	=
=
=	54.0 g·mol-1
96.0 g·mol-1
192.0 g·mol-1	
Al2(SO4)3	=	342.0 g·mol-1	

		We will need to find the molar mass of Al2(SO4)3:

		Calculate the concentration of Al2(SO4)3:

		Write a balanced equation for the dissociation reaction: 

				Al2(SO4)3 → 2 Al3+(aq) + 3 SO42-(aq)

		Using the balanced equation, calculate the concentration of the individual ions: 

			[Al3+] = 2× [Al2(SO4)3] = 2 × 0.0249 = 0.0498 M or 4.98 ×10-2M

			[SO42-] = 3× [Al2(SO4)3] = 3 × 0.0249 = 0.0747 M or 7.47×10-2M