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What is Q actually measuring?Write equation for precipitate formation:
2AgNO3(aq) + K2SO4(aq) --> Ag2SO4(s) +2KNO3(aq)
It is a Q type of calculation:
Ag2SO4(s) < -- > 2Ag+(aq) + SO4 2-(aq)
Q = [Ag+]^2 [SO4 2-]
Our goal to solve the question is to find these 2 concentrations so we can sub it into the Q
A dilution occurs when the two solutions are combined:
For AgNO3 conc
c1v1 = c2v2
(0.10)(100) = c2 (100+100)
c2(AgNO3) = 0.05 mol/L
As each AgNO3 has one Ag+ ion, we can say [Ag+] = 0.05 mol/L
Similarly, if we did this above steps for the K2SO4 solution we will get [SO4 2-] = 0.05 mol/L
Sub into Q:
Q = (0.05)^2 (0.05)
Q = 1.25 x 10^-4
Here Q > Ksp. This means there will be a shift to the left in the equilibrium (Ag2SO4(s) < -- > 2Ag+(aq) + SO4 2-(aq)). If we shift left that means a precipitate is made
Thank you so much!Write equation for precipitate formation:
2AgNO3(aq) + K2SO4(aq) --> Ag2SO4(s) +2KNO3(aq)
It is a Q type of calculation:
Ag2SO4(s) < -- > 2Ag+(aq) + SO4 2-(aq)
Q = [Ag+]^2 [SO4 2-]
Our goal to solve the question is to find these 2 concentrations so we can sub it into the Q
A dilution occurs when the two solutions are combined:
For AgNO3 conc
c1v1 = c2v2
(0.10)(100) = c2 (100+100)
c2(AgNO3) = 0.05 mol/L
As each AgNO3 has one Ag+ ion, we can say [Ag+] = 0.05 mol/L
Similarly, if we did this above steps for the K2SO4 solution we will get [SO4 2-] = 0.05 mol/L
Sub into Q:
Q = (0.05)^2 (0.05)
Q = 1.25 x 10^-4
Here Q > Ksp. This means there will be a shift to the left in the equilibrium (Ag2SO4(s) < -- > 2Ag+(aq) + SO4 2-(aq)). If we shift left that means a precipitate is made
I believe it's to do with the amount of dissolution. So Q essentially measures the same variable as Ksp, which is the concentration of products (gaseous or aqueous states) divided by the concentration of reactants, but it is at a specific point in time. So if Q is less than Ksp, this means the solution is unsaturated and more solid will continue to dissolve until the system reaches equilibrium, which occurs when Q = Ksp. Alternatively, if Q is greater than Ksp, then the solution is 'supersaturated', and so solid will precipitate until Q = Ksp (so reverse process). When Q = Ksp, the rate of dissolution is equal to the rate of precipitation, so there will be no net change in the amount of dissolved solid. Thus, there won't be a 'visible' precipitateWhat is Q actually measuring?
What is the method in calculating Q?
Also Why is it that if
Q > ksp = Precipitate
Q < ksp = no precipitate
Q = ksp = no precipitate but saturated?
Q is called the reaction quotient. It is used to describe the ratio of products to reactants in a reversible reaction. K (the equilibirum constant) is a specific type of Q, when a system is at equilibrium.What is Q actually measuring?
What is the method in calculating Q?
Also Why is it that if
Q > ksp = Precipitate
Q < ksp = no precipitate
Q = ksp = no precipitate but saturated?