A 100 L reaction container is charged with 0.724 mol of NOBr, which decomposes at a certain temperature** (say between 100 and 150 oC) according to the following reaction:
NOBr(g) ↔ NO(g) + 0.5Br2(g)
At equilibrium the bromine concentration is 1.82x10-3 M. Calculate Kc (in M0.5)
i got .067
At a certain temperature* (probably not 25 ºC), the solubility of silver sulfate, Ag2SO4, is 0.018 mol/L. Calculate its solubility product constant for this temperature. SIG. FIG. (required because number is small)
*Solubility product constants are very temperature sensitive. They are generally reported at 25 ºC. Not necessarily using this temperature allows me some flexibility.
i got 2.73e-5
At a certain temperature, the solubility of potassium iodate, KIO3, is 53.0 g/L. Calculate its solubility product constant for this temperature.
.059
Please help!
In the first problem, what is (M0.5)?
Ag2SO4 ==> 2Ag^+ + SO4^2-
0.018.....2*0.018..0.018
Ksp = (Ag^+)^2(SO4^2-)
Ksp = (2*0.018)^2*(0.018) = ? closer to 2.33E-5 I think.
KIO3 ==> K^+ + IO3^-
53.0/214 = about 0.248
Ksp = (K^+)(IO3^-)
Ksp = 0.248)(0.248) = ?
Frankly I think this is ridiculous. Does anyone use Ksp values for SOLUBLE materials?
thanks so much!
To calculate the solubility product constant (Ksp), we need two important pieces of information: the balanced equation for the dissolution of the substance and the solubility concentration.
For the first problem:
- We have the balanced equation: NOBr(g) ↔ NO(g) + 0.5Br2(g)
- The concentration of bromine (Br2) at equilibrium is given as 1.82x10^-3 M.
To calculate Kc, we need to express the concentration of each species in terms of Molarity (M). Since the equation does not provide the concentrations of NO(g) and NOBr(g), we'll need to use an ICE table to find their concentrations at equilibrium.
Let's assume 'x' M is the concentration of NO(g) formed and Br2(g) consumed at equilibrium. As the equation states, 1 mol of NOBr decomposes to form 0.5 mol of Br2, so the decrease in NOBr concentration due to decomposition is 0.5x.
The ICE table looks like this:
I 0.724 M 0 M 0 M
C -0.5x M +x M +0.5x M
E 0.724 - 0.5x M x M 0.5x M
To find the equilibrium concentration of Br2, we add the initial concentration (0 M) to the change in concentration at equilibrium (0.5x M):
0 + 0.5x = 1.82x10^-3 M
0.5x = 1.82x10^-3 M
x = (1.82x10^-3 M) / 0.5
x = 3.64x10^-3 M
Now that we have the concentration of Br2 at equilibrium, we can calculate Kc:
Kc = [NO][Br2^(0.5)] / [NOBr]
Kc = [x][(0.5x)^(0.5)] / [0.724 - 0.5x]
Kc = (3.64x10^-3 M)(0.5x^(0.5)) / (0.724 M - 0.5(3.64x10^-3 M))
Kc = 0.067 M^(0.5)
So, the value of Kc for the given reaction is 0.067 M^(0.5).
Now, moving on to the second problem:
- The solubility of silver sulfate (Ag2SO4) is given as 0.018 mol/L.
The balanced equation for the dissolution of Ag2SO4 is:
Ag2SO4(s) ↔ 2Ag+(aq) + SO4^2-(aq)
Since 2 moles of Ag+ ions are formed for every mole of Ag2SO4 dissolved, the concentration of Ag+ will be 2 times the solubility concentration:
[Ag+] = 2(0.018 mol/L) = 0.036 mol/L
Similarly, since 1 mole of SO4^2- ions are formed for every mole of Ag2SO4 dissolved, the concentration of SO4^2- will be equal to the solubility concentration:
[SO4^2-] = 0.018 mol/L
Now, we can write the expression for the solubility product constant, Ksp:
Ksp = [Ag+]^2[SO4^2-]
Ksp = (0.036 mol/L)^2(0.018 mol/L)
Ksp = 2.73x10^-5 mol^3/L^3
So, the solubility product constant (Ksp) for silver sulfate at the given temperature is 2.73x10^-5 mol^3/L^3.
Finally, for the third problem:
- The solubility of potassium iodate (KIO3) is given as 53.0 g/L.
To calculate the solubility product constant (Ksp), we need to convert the solubility from grams per liter to moles per liter:
Molar mass of KIO3 = 214 g/mol
53.0 g/L ÷ 214 g/mol = 0.248 mol/L
Since the balanced equation for the dissolution of KIO3 is:
KIO3(s) ↔ K+(aq) + IO3^-(aq)
We can deduce that the concentration of K+ is equal to the solubility concentration:
[K+] = 0.248 mol/L
Similarly, the concentration of IO3^- is also equal to the solubility concentration:
[IO3^-] = 0.248 mol/L
Now, we can write the expression for Ksp:
Ksp = [K+]^[IO3^-]
Ksp = (0.248 mol/L)^(0.248 mol/L)
Ksp = 0.059 mol
So, the solubility product constant (Ksp) for potassium iodate at the given temperature is 0.059 mol.