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First Order Second Order Al 693 kIA] Directions: Choose the ONE BEST answer for

ID: 1045469 • Letter: F

Question


First Order Second Order Al 693 kIA] Directions: Choose the ONE BEST answer for each question 1. The slightly soluble salt barium arsenate, Ba,(AsO),[Molar Mass-689.83], dissolves in water as follows: Ba, (Aso)3Ba +2 Aso, The solubility of Ba, (Aso)2(s) in pure water is 2.6 x10 * grams per liter (g/L) Calculate the value of Kp (Note: this compound is not found in your K table.) a. 1.4x1021 b. 7.6x 1053 c. 8.2 x10s? d. 8.3x1021 e. 1.3x10" 2. The K of bismuth iodide, Bil, Molar Mass-589.68], is 8.1 x 10. If Bil, (s) is dissolved in pure water, what is the solubility in grams per liter (g/L)? (Suggestion: Write the dissolving equation and K, expression for Bil, before starting this problem.) a. 3.1 x 10g/L b. 7.8x10 g/L c. 1.0x10 g/L d 1.3 x10'g/L e. 1.8x 10 /L 3. what is the solubility of Bil, (s) (in g/L) in a solution that contains 0.01 M Bi"(aq)? a 2.6 x 10 g/L b. 8.5x 10 g/L c. 2.7x10"g/L d 1.4x 10 g/L e. 4.8x10 g/

Explanation / Answer

1 . Option C i.e 8.2 x 10-51

Explanation :-

because the unit of solubility products (Ksp) always considered in mol/L , therefore convert the units of solubility from g/L to mol/L by using the formula :

number of moles = given mass in gram / gram molar mass

i.e number of moles of Ba3(AsO4)2 = 2.6 x 10-8 g / 689.83 g /mol

= 3.769 x 10-11

Therefore , solubility of Ba3(AsO4)2 = 3.769 x 10-11 mol / L

now , given Ba3(AsO4)2 <-----------> 3Ba2+ + 2AsO43-

therefore concentration of Ba2+ = [Ba2+] = 3 x 3.769 x 10-11 = 11.307 x 10-11 M

also , concentration of AsO43- = [ASO43-] = 2 x 3.769 x 10-11 = 7.538 x 10-11 M

Now , Expression of Ksp for the given equilibrium reaction = [Ba2+]3   [AsO43-]2

= ( 11.307 x 10-11)3 ( 7.538 x 10-11)2

= 1445.58 x 10-33 x 56.821 x 10-22

= 8.2 x 10-51

Hence Ksp value is 8.2 x 10-51 i.e Option (c) is correcr answer .

note - Solubility Product of sprangly soluble salt is equal to the product of molar concentration of products, raised to power their stoichiometric coefficients at equilibrium stage .