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!
3 answers
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?
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!
1 answer