Chemistry, asked by maazshaikh4493, 7 months ago

4. Calculate the normality of Mg(OH)2 solution that contains 119.35 mg
Mg(OH)2 per 0.10 L of solution.

Answers

Answered by Malkarajpoot24
1

Answer:

normality = mass/equavelent mass x 1000/volume(ml)

0.11935/29.2 x 1000/100 ml

normality = 0.0408

Explanation:

Answered by uttaratejas
0
The pH of Mg(OH)
2

solution is 10.45 at 25

C . The solubility product of magnesium hydroxide will be:
A
2.24×10
−11
M
3

B
1.12×10
−11
M
3

C
3.36×10
−11
M
3

D
5.60×10
−12
M
3

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ANSWER
The pH of the magnesium hydroxide solution is 10.45

Hence pOH=14−10.45

=3.55

[OH

]=antilog(−3.55)

=0.000282 M.

[Mg
2+
]=
2
0.000282



=0.000141 M.

The expression for the solubility product is K
sp

=[Mg
2+
][OH

]
2


=0.000141×(0.000282)
2


=1.12×10
−11
M
3
.

Option B is correct.
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