destinyranson
23.03.2020 •
Chemistry
A saturated solution of Cu(IO3)2 was prepared in 0.0250 M KIO3. The copper concentration was found to be 0.00072 M. Set up an ICE chart and calculate the equilibrium (IO3-) and the Ksp
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Ответ:
0.02644 M = [IO3-]
Ksp = 0.0000005 = 5*10^-7
Explanation:
Step 1: Data given
Asaturated solution of Cu(IO3)2 was prepared in KIO3
Concentration of KIO3 = 0.0250 M
The copper concentration was found to be 0.00072 M
Step 2: Equations
Cu(IO3)2(s) + aq ⇔Cu2+(aq) + 2IO3 - (aq)
Step 3: The initial concentrations
The initial concentration of KIO3 = 0.0250 M
The initial concentration of Cu2+ is 0 M
Step 4: The changed concentrations
After preparing the Cu2+ in KIO3-; the concentration of Cu2+ will be changed with X
For every 1 Cu2+ we have 2 IO3-
This means if the concentration of Cu2+ is changed by X; the concentration of IO3- is changed by 2X. This gives a total concentration of 0.0250 + 2X for IO3-
Step 4:
Said is that the copper concentration was found to be 0.00072 M
This means the final concentration of Cu2+ is 0.00072 M or also said X = 0.00072
Now we can calculate the final IO3- concentration
0.0250 + 2X = [IO3-]
0.0250 + 2* 0.00072 = [IO3-]
0.0250 + 0.00144 = [IO3-]
0.02644 M = [IO3-]
Step 5: Calculate Ksp
Cu(IO3)2(s) ⇔Cu2++ 2IO3 -
Ksp of Cu(IO3)2 = [Cu+2]*[IO3-]²
Ksp = 0.00072 * 0.02644²
Ksp = 0.0000005 = 5*10^-7
Read also: link
Ответ:
Explanation:
1. The ICE table
2. Concentration of IO₃⁻
At equilibrium, [Cu²⁺] = 0.000 72 mol·L⁻¹, so x = 0.000 72.
The new ICE table becomes
3. Ksp
Ответ:
The correct answer is C
Explanation: I took the test