MolSynapse
question bank
All chapters
NEET
JEE
Practice
MolSynapse
/
Question bank
/
Ionic Equilibrium I
/
Solubility product and simple solubility calculations
JEE Main level
Numerical answer
Difficulty 2 of 3
ie191
K
s
p
of
BaSO
X
4
is
1.2
×
1
0
−
10
(molar mass
233
g
mo
l
−
1
). Calculate the mass of
BaSO
X
4
that dissolves in
500
mL
of pure water, in mg (two decimals).
Numerical answer type, answer in mg, tolerance 0.02.
Answer: 1.28 mg
s
=
K
s
p
=
1.1
×
1
0
−
5
M
. Mass
=
s
×
233
×
0.5
L
=
1.276
e
−
03
g
=
1.28
mg
.
Practice Ionic Equilibrium I
More on Solubility product and simple solubility calculations
Previous question
Next question
More questions on Solubility product and simple solubility calculations
The solubility of PbCl2 (molar mass 278 g mol^-1) in water is 4.17 g L^-1. Its K_sp is approximately
JEE Main level
The K_sp of Ag2CrO4 is 1.5 x 10^-12. Calculate its molar solubility in pure water, expressed in units of 10^-5 mol L^-1 (two decimals).
JEE Main level
K_sp of Ca(OH)2 is 4.0 x 10^-6. Taking log 2 = 0.301, calculate the pH of its saturated solution at 298 K (two decimals).
JEE Main level
K_sp of AgBr is 5.0 x 10^-13 (molar mass 188 g mol^-1). What volume of water, in litres, is needed to dissolve 10.0 mg of AgBr? (nearest int...
JEE Main level
For a saturated solution of Ag2CrO4 (K_sp = 1.1 x 10^-12) at 298 K, choose the correct statement(s).
JEE Main level
Three sparingly soluble salts AB, AB2 and AB3 each have molar solubility 1.0 x 10^-4 mol L^-1. Choose the correct statement(s).
JEE Main level