Question

A particular water sample has 131 ppm of CaSO4 (131 g CaSO4 per 10^6 g water) if this water is boiled in a teakettle , aproximate what fraction of the water must be evaporated before CaSO4(s) begins to precipitate? (Ksp CaSO4=9.1x10^-6)

Answer 1

Hey dear,

● Answer-
0.68

● Explaination-
# Given-
C = 131 ppm
Ksp = 9.1×10^-6

# Solution-
For the reaction,
CaSO4 ---> Ca2+ + (SO4)2-
Let s be solubility.
Ksp = s^2
9.1×10^-6 = s^2
s = 3×10^-3

We also have, 131 g of CaSO4 in 10^6 g of water.
No of moles of CaSO4 in 10^6 g water = 131/136 = 0.9632

Let l ml be amount of water when CaSO4 precipates in it.
M = 0.9632 / (x/1000)10^6
M = 9.632×10^-4 / x

That is -
M = s
9.632×10^-4 / x = 3×10^-3
x = 0.32

Water to be evaporated is
V' = (1-x)V
V' = (1-0.32)V
V' = 0.68V

Therefore, 0.68 fraction of the water is to be evoporated.

Hope this helps...

Answer 2

"Given,

Water sample contains 131 ppm of ${ CaSO }_{ 4 }$

Where,

131g of ${ CaSO }_{ 4 } in { 10 }^{ 6 }$of water

${ K }_{ sp } of { CaSO }_{ 4 } is 9.1\quad \times \quad { 10 }^{ -6 }$

Consider the equation${ CaSO }_{ 4 }\quad \rightarrow \quad { Ca }^{ 2+ }\quad +\quad { \left( { SO }_{4}\right)}^{2-}$

Let 's' be solubility.

${ K }_{ sp }\quad =\quad { s }^{ 2 }$

$9.1\quad \times \quad { 10 }^{ -6 }\quad =\quad { s }^{ 2 }$

$s\quad =\quad 3\quad \times \quad { 10 }^{ -3 }$

We also have, 131 g of ${ CaSO }_{ 4 } in { 10 }^{ 6 } g$ of water.

No of moles of ${ CaSO }_{ 4 } in { 10 }^{ 6 } g$ water = 131/136 = 0.9632

Let l ml be amount of water when CaSO4 precipitates in it.

$M\quad =\quad \frac { 0.9632 }{ \left( \frac {x}{1000}\right) \quad \times \quad {10}^{6}}$

$M\quad =\quad \frac { 9.632\quad \times \quad { 10 }^{ -4 } }{ x }$

That is $-M\quad =\quad s$

$\frac { 9.632\quad \times \quad { 10 }^{ -4 } }{ x } \quad =\quad 3\quad \times \quad { 10 }^{ -3 }$

$x\quad =\quad 0.32$

Water to be evaporated is

$V'\quad =\quad \left( 1\quad -\quad x \right) V$

$V'\quad =\quad \left( 1\quad -\quad 0.32 \right) V$

$V'\quad =\quad 0.68V$

Therefore, 0.68 fraction of the water is to be evaporated.

Percentage$\quad =\quad \left( \frac { 68 }{100}\right) \quad \times \quad 100$

In percentage around 68% of water is to be evaporated."