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.
Mg(OH)2 + H2O-à Mg+2 + 2 OH-
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
Help with questions 3 and 4 please!
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The fully balanced equation can be withdrawn from those trials where the reaction has taken place completely and none of the reactant is left in the last table.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
Table 1. The results of adding solid Mg(OH)2 to 10.0L of water. Total amount of Mg(OH)2...
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) [Mg] in solution (M) (OH) in solution (M) Mass of Mg(OH)2 that does NOT dissolve (g) 0.00963 0.04815 0.09590 0.09630 0.09700 0.10000 0.15000 0.20000 1.65 x 104 8.26 x 104 1.64 x 10 1.65 x 10 1.66 x 10 1.71 x 10 2.57 x 103 3.43 x 10 1.65x 10 8.26 x 10 1.64 x 10...
Table 1. The results of adding solid Mg(OH)2 to 10.0L of water. Total amount of Mg(OH)2 added (g) [Mg] in solution (M) [OH') in solution (M) Mass of Mg(OH)2 that does NOT dissolve (g) 0.00963 0.04815 0.09590 0.09630 0.09700 0.10000 0.15000 0.20000 Total amount of Mg(OH)2 added (mol) 1.65 x 104 8.26 x 10" 1.64 x 100 1.65 x 10 1.66 x 103 1.71 x 10 2.57 x 10 3.43 x 100 1.65x 10 8.26 x 10 1.64 x 104...
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