Question

How many grams of calcium hydroxide are
required to prepare 10-2 M and 10-2 N solution?
​

Answer 1

Explanation:

since dissociation is complete, hence, α=1

Now, we know that,

pH+pOH=14

pOH=14−pH=14−10.65=3.35

pOH=−log

10

​

[OH

−

]

[OH

−

]=Antilog(−3.35) or 10

−3.35

=4.46×10

−4

mol/dm

3

Now,

[OH

−

]=nαC

where, n=Acidity of the given base or no. of replaceable OH

−

 ion=2

Hence,

C=

2×1

4.46×10

−4

​

=2.23×10

−4

We know that

C= Molarity of the solution=

Volume of solution in dm

3

No. of moles of solute

​

Hence, no. of moles of Ca(OH)

2

​

=2.23×10

−4

×0.25=5.57×10

−5

mol