Home
Study
Practice
Resources
Download Offline
Chemical Equilibrium Examples
Equilibrium Constants
Example
Solubility Product Constant (
K
s
p
K_{sp}
K
s
p
) of Gypsum
Calculate the solubility of Calcium Sulfate (
C
a
S
O
4
CaSO_4
C
a
S
O
4
, gypsum) in pure water at
25
∘
C
25^\circ\text{C}
2
5
∘
C
if its
K
s
p
K_{sp}
K
s
p
is
4.93
×
10
−
5
4.93 \times 10^{-5}
4.93
×
1
0
−
5
. Express the answer in
g/L
\text{g/L}
g/L
(Molar mass
=
136.14
g/mol
= 136.14 \, \text{g/mol}
=
136.14
g/mol
).
Step-by-Step Solution
0 of 4 Steps Completed
Show All
1
Start Solution
Click to reveal the first step