UNIT 8: Ionic Equilibrium - Online Test
pH = -log10[H+]
∴[H+]=10-pH
Let the volume be x mL
V1M1 + V2M2
+ V3M3 =VM
∴
x mL of 10-1M+ x mL of
10-2M + x mL of 10-3 M
= 3x mL of [H+]
∴[H+ ] = x[0.1+0.01+0.001] / 3x
= ( 0.1+ 0.01+ 0.001 ) / 3
= 0..111 / 3
= 0.037
= 3.7 ×10−2
AgCl(s) ↔ Ag+ (aq) + Cl-(aq)
NaCl (0.1 M) → Na+(0.1
M) + Cl- (0.1 M)
Ksp = 1..6 ×10-10
Ksp = [Ag+][Cl-]
Ksp = (s)(s+0.1)
0.1>>>s
∴
s + 0.1 = 0.1
∴
S = [ 1..6 ×10-10 ] / 0.1 = 1.6 ×10-9
PbI2 (s) ↔ Pb2+
(aq) + 2I- (aq)
Ksp = (s)(2s)2
3.2 × 10-8 = 4s3
s =
(3.2×10-8 / 4)1/3
= (8 × 10-9 )1/3
= 2 × 10-3M
= 2 × 10-3M
Addition of salt KY (having a
common ion Y–) decreases the solubility of MY and NY due to common ion effect.
Option (a) and (b) are wrong.
For salt MY , MY ↔ M+ + Y-
Ksp = (s)(s)
6.2 × 10-13 = s2
∴ s= √(6.2 × 10-13) = 10-7
for salt NY3 ,
NY3 ↔ N3+
+ 3Y-
Ksp = (s )(3s)3
Ksp = 27s4
s = [ (6.2 ×10−13)/27 ]1/4
s = 10-4
The molar solubility of MY in water is less
than of NY3
x ml of 0.1 M NaOH + x ml of 0.01 M HCl
No of moles of NaOH = 0.1 × x × 10–3 = 0.1x
× 10-3
No of moles of HCl = 0.01 × x × 10-3 = 0.01x × 10-3
No of moles of NaOH after mixing
= 0.1x × 10-3 - 0.01x × 10-3
= 0.09x
× 10-3
Concentration of NaOH= [0.09x × 10−3 ] /
[2x × 10−3] = 0.045
[OH-] = 0.045
POH = −log (4.5 × 10−2
)
= 2 −log 4.5
= 2- 0.65 = 1.35
pH = 14-1.35=12.65
K a =1 ×10-3
pH=4
[salt] /
[Acid] = ?
pH = pKa +log {[Salt]/[Acid]}
4 = - log10 (1 ×10-3 ) + log {[Salt]/ [Acid]}
4 = 3 + log ([Salt]/ [Acid])
1= log10 ([Salt]/[Acid])
[Salt] / [Acid] =101
i.e., [Salt] / [Acid] = 1 / 10
1:10
KOH →
K+ + OH-
10-5m 10-5m 10-5m
KOH (10-5m) → K+
(10-5m) + OH- (10-5m)
[OH-] = 10-5M.
pH= 14 - pOH
pH = 14 - ( - log [OH-]
)
= 14 + log [OH- ]
= 14 + log10−5
= 14−5
= 9.
H3PO4
+H − OH ↔ H3O+ + H2PO4-
acid 1 +
base 1 ↔ acid 2 + base 2
∴H2PO4− is the conjugate base of H3PO4
HPO42− can have the ability to accept a proton to form H2PO−4.
It can also have the ability to
donate a proton to form PO4-3
pH = -log10[H+
]
∴[H+ ]=10-pH
=100
=1
[H+] = 1M
The solution is strongly acidic
