Question

Table 1. The results of adding solid Mg(OH)₂ to 10.0L of water.

Total amount of Mg(OH)2 added (g)	Total amount of Mg(OH)2 added (mol)	[Mg2+] in solution (M)	[OH-] in solution (M)	Mass of Mg(OH)2 that does NOT dissolve (g)
0.00963	1.65 x 10-4	1.65x 10-5	3.30 x 10-5	0
0.04815	8.26 x 10-4	8.26 x 10-5	1.65 x 10-4	0
0.09590	1.64 x 10-3	1.64 x 10-4	3.29 x 10-4	0
0.09630	1.65 x 10-3	1.65 x 10-4	3.30 x 10-4	0
0.09700	1.66 x 10-3	1.65 x 10-4	3.30 x 10-4	0.0007
0.10000	1.71 x 10-3	1.65 x 10-4	3.30 x 10-4	0.0037
0.15000	2.57 x 10-3	1.65 x 10-4	3.30 x 10-4	0.0537
0.20000	3.43 x 10-3	1.65 x 10-4	3.30 x 10-4	0.1037

Answer the following questions using Table 1 and your textbook.

1. Write the fully balanced chemical equation for Mg(OH)₂ dissolving in water.

Mg(OH)₂ + H₂O-à Mg⁺² + 2 OH^-

2. When 8.26 x 10⁻⁴ moles of Mg(OH)₂ are added, a. Why is [Mg(OH)₂] = 8.26 x 10⁻⁵ M? b. Why is [Mg²⁺] = 8.26 x 10⁻⁵ M? c. Why is [OH⁻] = 1.65 x 10⁻⁴ M?

a.)/b.) concentration of Mg(OH)² = 8.26x10^-4 mol /10L = 8.26 x 10^-5M. Thus Mg²⁺ = 8.26x10^-5M

c.) 8.26 x 10-5 x 2 = 1.65 x 10-4

3. When 0.20000g Mg(OH)₂ are added, how many grams dissolve? (Show your work.)

4. When 0.09700g Mg(OH)₂ are added, how many grams dissolve? (Show your work.)

Help with questions 3 and 4 please!

Answer 1

1-

The fully balanced equation can be withdrawn from those trials where the reaction has taken place completely and none of the reactant is left in the last table.

Ex-

Trial 1 1.65 x 10-4 moles  Mg(OH)2 + H2O ----------> 1.65 x 10-4 mole Mg+2 + 3.30 x 10-5 moles OH-

Now if we devide alll by the common facotor 1.65 x 10-4 , ,then we will get

  Mg(OH)2 + H2O ----------> Mg+2 + 2 OH-

So the balance reaction with the least possible co-efficient is

  Mg(OH)2 + H2O ----------> Mg+2 + 2 OH-

i.e 1 mole of Mg(OH)2 dissociate into one mole of Mg+2 and 2 moles of OH- ions

2-

Now given moles of Mg(OH)2 taken = 8.26 x 10−4 moles

Now we have total volume of solution = 10.0L

Again we know Molarity = moles of solute / volume of solution in L

= 8.26 x 10−4 moles / 10.0L

= 8.26 x 10−5 moles /L

= 8.26 x 10−5 M

3-

That means the concentration of Mg(OH)2 taken = 8.26 x 10−5 M

So following the balanced chemical equation after complete reaction -

concentration of Mg+2 formed  = same as that of concentration of Mg(OH)2 taken = 8.26 x 10−5 M

concentration of OH- formed  = 2 * concentration of Mg(OH)2 taken = 8.26 x 10−5 M * 2 = 1.65 x 10−4 M

4-

Given mass of  Mg(OH)2 taken = 0.20000 g

Now in this trial,

given concentration of Mg+2 formed = 1.65 x 10-4 M

That means moles of Mg+2 formed = concnetration * volume

= 1.65 x 10-4 M* 10 L

= 1.65 x 10-4 mole / L* 10 L

= 1.65 x 10-3 mole

That means moles of Mg(OH)2 dissolved = 1.65 x 10-3 moles

Or we can say

Mass of Mg(OH)2 dissolved = moles * molar mass of Mg(OH)2

= 1.65 x 10-3 moles * 58 g/mole

= 0.0957‬ g

5-

Given mass of  Mg(OH)2 taken = 0.09700 g

Similarly, in this trial, moles of Mg+2 formed = 1.65 x 10-3 moles

That means moles of Mg(OH)2 reacted or dissolved = 1.65 x 10-3 moles

Or we can say

Mass of Mg(OH)2 dissolved = moles * molar mass of Mg(OH)2

= 1.65 x 10-3 moles * 58 g/mole

= 0.0957‬ g