CHEM 104 Module 3 Exam – Portage Learning 2026 – All Questions with Verified Solutions

M3: Exam
Question 1
pts

Click this link to access the Periodic Table. This may be helpful throughout the exam.



Show the calculation of the molar solubility (mol/L) of Mg(OH)2, Ksp of Mg(OH)2 = 1.8 x 10-11.

Your Answer:

ksp= [Mg+2] [OH−] 2



= (s ) (2s) 2



s 3 = 1.8 x 10 −



s= 3 ( 1.8 x 10 − 11 4 )



s= 1.65 x 10 − 4 mol/L




Mg(OH)2 (s) Mg+2 (aq) + 2 OH-1 (aq)

(s) (s) (2s)



1.8 x 10-11 = [Mg+2] x [OH-1]2

1.8 x 10-11 = [s] x [2s]2

, 1.8 x 10-11 = 4 s3

s = 1.65 x 10-4 mol/L

Format exponents correctly.

Question 2
pts

Click this link to access the Periodic Table. This may be helpful throughout the exam.



Show the calculation of the Ksp of MnS if the solubility of MnS is 0.0001375 g/100 ml.

MW of MnS = 87.01

MnS (s) Mn+2 (aq) + S-2 (aq)

Your Answer:

mol sol MnS= ( 0.0001375 g/100ml) x (1000ml/1L) x (1 mol/87.01 g)

mol sol MnS= 1.375 x 10 − 6 x 1000 x .01149



mol sol= 1.58 x 10 – 5 mol/L



ksp= (1.58 x 10 – 5 ) (1.58 x 10 – 5 )= 2.50 x 10 – 10



molar sol of MnS = (0.0001375 g/100 ml) x (1000 ml / 1 L) x (1 mol / 87.01 g MnS)

molar sol MnS = 1.58 x 10-5 mol / L

This means that as 1.58 x 10-5 mol / L of MnS dissolves, 1.58 x 10-5 mol / L of Mn+2 and 1.58 x
10-5 mol / L of S-2 are formed since one mole of each ion forms from 1 mole of MnS



Ksp = [Mn=2] x [S-2] = (1.58 x 10-5) x (1.58 x 10-5) = 2.50 x 10-10