\"Hard\" water contains alkaline earth cations such as Ca^2+, which reacts with
ID: 1023159 • Letter: #
Question
"Hard" water contains alkaline earth cations such as Ca^2+, which reacts with CO_3^2- to form insoluble deposits of CaCO_3. You may want to reference (pages 700 - 704) Section 16.15 while completing this problem. What is the ion product when a 250 mL sample of hard water having [Ca^2+] = 8.0 Times 10^-4 M is treated with a 0.10 mL sample of 2.0 Times 10^-3 M Na_2 CO_3? Express the ion product to two significant digits. What is the ion product when a 250 mL sample of hard water having [Ca^2+] = 8.0 Times 10^-4 M is treated with a 0.10 mL sample of 10 mg of solid Na_2 CO_3? Express the ion product to two significant digits.Explanation / Answer
Let us find the ionic product of part A
Total volume = 250 ml +0.1ml = 250.1 ml
Moles of CO32- = 2.0*10^-3 moles / 1.0 L* 1.0*10^-4 L
= 2.0*10^-7 moles
New Molarity of CO32- = 2.0*10^-7 /0.2501
= 7.996*10^-7
Moles of Ca2+ = 8.0*10^-4 moles / L*0.250 L
= 2.0*10^-4 moles
New Molarity o f Ca2+ = 2.0*10^-4 moles /0.2501
= 7.996*10^-4
Therefore the ionic product of CaCO3 is
[Ca2+] [CO32-]
= [7.996*10^-4 ][ 7.996*10^-7 ]
= 6.394*10^-10
Let us find the ionic product of part B
Number of moles = 0.01 g/ 105.988 g/ moles
=9.44 *10^-5 moles
Molarity = number of mole s/ Volume in L
Here volume = 0.10 ml = 1.0*10^-4 L
== 9.43 *10^-5 moles/ 1*10^-4
= 0.943 M
Moles of Na2CO3 or CO3 = 0.943 moles / L*10^-4 L
= 9.430*10^-5 Moles
Molar mass of Na2CO3 = 105.9888 g/mol
Total volume = 250 +.1 = 250.1 ml
New Molarity of CO32- = 9.430*10^-5 /0.2501
= 3.77 *10^-4
Moles of Ca2+ = 8.0*10^-4 moles / L*0.250 L
= 2.0*10^-4 moles
New Molarity o f Ca2+ = 2.0*10^-4 moles /0.2501
= 7.996*10^-4
Therefore the ionic product of CaCO3 is
[Ca2+] [CO32-]
= [7.996*10^-4 ][ 3.77 *10^-4]
= 3.0*10^-7
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