CHEM 2 0 1 Self Qui z - 3 (Eq u i l i bri um cal cu l at i o ns)
1. Write a charge balance for a solution containing H+, OH -, Ca2+, HCO3 -, CO3 2-,
Ca(HCO3)+, Ca(OH)+, K+, ClO4 -.
[H+] + 2[Ca2+] + [Ca(HCO3)+] + [Ca(OH)+] + [K+] =
= [OH -] + [HCO3 -] + 2 [CO3 2-] + [ClO4 -]
2. Considering the equilibria below, derive an equation relating [M2+], K1 and K2 when
0.10 mol of MX2 is dissolved in 1 L of water.
M 2+ + X ! " MX +
K1 =
MX + + X ! " MX 2 (aq )
K2 =
[MX + ]
[M 2+ ][ X ! ]
[MX 2(aq ) ]
[MX + ][ X ! ]
[MX + ] = K1[M 2+ ][ X ! ]
[MX 2(aq ) ] = K 2 [MX + ][ X ! ] = K 2K1[M 2+ ][ X ! ]2
Mas s balance equation:
MX 2 =
0.10 mol
= 0.10 M = [MX 2(aq ) ] + [MX + ] + [M 2+ ] =
1L
= K 2K1[M 2+ ][ X ! ]2 + K1[M 2+ ][ X ! ] + [M 2+ ]
Charge balance equation:
2 [M 2+ ] + [MX + ] = [ X ! ]
By substituting of [M2+] and [MX-] we have an expression for the [X-]
2 [M 2+ ] + K1[M 2+ ][ X ! ] = [ X ! ]
By solving this equation for [X-], we have:
[X ! ] =
2[M 2+ ]
1! K1[M 2+ ]
Now we put expression for [X-] into the mass balance equation:
" 2[M 2+ ] % 2
2[M 2+ ]
2+
0.10 M = K 2K1[M ]$
+
K
[
M
]
+ [M 2+ ]
'
1
2+
2+
1!
K
[
M
]
1!
K
[
M
]
#
&
1
1
2+
3. When ammonium sulfate dissolves, both the anion and cation have acid-base
reactions:
+
(NH 4 ) 2SO 4 ! 2NH 4 + SO 4
+
2"
K sp = 276
+
K a = 5.7 "10#10
K b = 9.8 #10!13
NH 4 ! NH 3 (aq ) + H
SO 4 2! + H 2O " HSO 4 ! + OH !
a. write a charge balance for this system
[NH 4 + ] + [ H + ] = 2 [ S O 4 2- ] + [ HS O 4 - ] + [ O H - ]
b. write a mass balance for this system
[NH 3 ] + [ NH 4 + ] = 2
{[ S O
24
] + [ HS O 4 - ]
}
c. find the concentration of the NH3(aq) if the pH is 9.25
[H+] = 10-9.25 and [OH-] = 10-4.75
[NH 3 ][H + ]
[NH 3 ]!10"9.25
"10
=
5.7
!10
=
+
+
[NH 4 ]
[NH 4 ]
#
[NH 3 ]
"0.75
= 1.014
+ = 10
[NH 4 ]
Therefore,
[NH3 ] = 1.014 [NH4 +]
From another hand,
[HSO 4 ! ][OH ! ]
[HSO 4 ! ]10!4.75
!13
= 9.8 "10 =
2!
2!
[SO 4 ]
[SO 4 ]
[HSO 4 ! ]
#
= 9.8 "10!8.25 = 5.51"10!8
2!
[SO 4 ]
Therefore,
2!
!
[HSO 4 ] = 5.51"10!8 [SO 4 ]
Putting these values of [NH3] and [HSO4 -] into the mass balance gives:
+
+
+
2"
(
2"
2"
)
1.014 ![NH 4 ] + [NH 4 ] = 2 [SO 4 ] + 5.51!10"8 [SO 4 ]
1.007 ![NH 4 ] = [SO 4 ]
Now an expression for the Ksp is used:
K sp = [NH 4 + ]2 [SO 4 2! ] = 1.007 [NH 4 + ]3 " [NH 4 + ] = 3
+
[NH 3 ] = 1.014[NH 4 ] = 1.014 ! 6.5 = 6.59M
276
= 6.5M
1.007