So here we continue our discussion of the comparison of the reaction quotient q to any type of equilibrium constant including Ksp. Now in these next two scenarios, they will not be equal to one another. Because q and k will not be equal to one another, this will cause my chemical reaction to shift either in the forward direction or the reverse direction to reestablish equilibrium. Now in this first example here, we're given new concentrations for our ions. Here we have 0.250 molar for each one and when we plug them into the equilibrium expression for Q, we can see that the value of Q comes out to 6.25 × 10-2.
In this case, we're going to see that Q is bigger than our Ksp, which is still the same value. Remember, the only way this Ksp or any type of k would change is if there were a change in temperature. Our Ksp remains the same because the temperature is being held constant, but the concentration of my ions are changing which will have a direct impact on the value of q. Here, q is larger than Ksp. If q is larger than Ksp or k in general, then our reaction will shift.
So, in terms of our number line, q will always shift to wherever k is. Here, q will shift this way to get to k. And if it shifts that way to get to k, then our chemical reaction will also shift as a whole to get to equilibrium. Remember, wherever you shift, that direction will cause an increase. Here we're shifting towards the reactant side, so the reactant side would be increasing.
If the reactant side is increasing, that means our product side is decreasing. We're going to say, If q is larger than k, then our chemical reaction will shift to the reactant side. In this case, because we're shifting towards the reactant which happens to be a solid, that means that the solubility will increase. Now in this case, we have new concentrations yet again for the ions. Again, the temperature is not being affected, so our equilibrium constant Ksp remains the same.
In this case, we have these new concentrations which give us a new q value which is 5.06 × 10-24. We can now see that value is less than Ksp. So if q is smaller than KSP or k in general, so now q is no longer here. Instead, q will be over here. It's going to be smaller.
So we're going to have to shift this way now. We're going to shift that way in terms of the number line and in terms of the chemical reaction. Remember, wherever we're heading towards, that side is increasing. This side would increase. This side here would decrease.
Our chemical reaction will shift to the product side. That means that the solubility of my ionic compound is increasing because I'm making more ions, and the amount of precipitated solid will decrease. Just remember, q, our reaction quotient is just used as a way of determining if we are at equilibrium. When q equals Ksp, we're at equilibrium. We will say that our chemical reaction or actually our solution is a saturated solution with no real dominant movement to the reactant side or the product side.
We're at equilibrium. Everything is balanced, so we'd represent a saturated solution. Then if our q happens to be larger than our Ksp, that means we have an excess of ions, so that's why the reaction is shifting in the reverse direction towards the reactants to recreate some more precipitate. We'd say that this type of process happens within a supersaturated solution. And then finally, when q is less than Ksp, we can continue to break down our ionic solid to produce more ions.
When q is less than Ksp, we'd say that we are in an unsaturated solution. Now that we've explored the connection between our reaction quotient, q, and Ksp, we'll be more prepared when we take a look at questions dealing with more calculation-based questions on the formation or lack thereof of precipitates.