Answer:
[A
g
+
] = 3.12 *10^-6 M
Explanation:
Step 1: Data given
Volume silver nitrate = 369 mL
Molarity silver nitrate = 0.373 M
Volume sodium chromate = 411 mL
Molarity sodium chromate = 0.401 M
The Ksp of silver chromate is 1.2 * 10^− 12
Step 2: The balanced equation
2
A
g
+
(
a
q
) + C
rO4^2- (
a
q
) →Ag2CrO4
( s)
Step 3:
[Ag+]i = [AgNO3] * V1/(V1+V2) * 1 mol Ag+ / 1 mol AgNO3
[Ag+]i = 0.373 M * 0.369/ (0.369+0.411)
[Ag+]i = 0.176 M
[C
rO4^2]i = [Ag2CrO4] * V2 /(V1+V2) * 1mol C
rO4^2 / 1 mol Ag2CrO4
[C
rO4^2i = 0.401 M * 0.411 / (0.369+0.411)
[C
rO4^2]i = 0.211 M
Ksp = [Ag+]²[CrO4^2-] = 1.2 * 10^− 12
[CrO4^2-]f = [CrO4^2-]i - 0.5 * [Ag+]i
[CrO4^2-]f = 0.211 -0.088 = 0.123 M
1.2 * 10^− 12 = (
2
x
)
²*(
0.123M
+
x
)
[
C
O
3^
−2]
f >> x
1.2 * 10^− 12 = (
2
x
)
²*
0.123M
x = 1.56 * 10^-6
[A
g
+
]
f = 2x = 3.12 *10^-6 M
,
