# Solutions and Precipitates [closed]

If a solution of $$\pu{20mL}$$ of $$\pu{ 0.050M}~\ce{K+ }$$is added to $$\pu{80mL}$$ of $$\pu{0.50M}\ce{ClO4-}$$ will a precipitate form and what is the value of $$Q_\mathrm{sp}$$? For $$\ce{KClO4}, K_\mathrm{sp}= 1.07 \times{10^{-2}}$$

*I know that we have to show what we tried originally, which is great and I think the best way to learn. I am honestly looking at this problem and have no idea what any of it means. I'm self-teaching chemistry from a textbook, and the sample problems were nothing like this! I looked online for videos but I don't know what I am looking for, and I am unsure of the difference between $$Q_\mathrm{sp}$$ and $$K_{\mathrm sp}$$?? Does anyone have some kind of video, or sample problem, or explanations and definitions that could help me solve this on my own? I don't know what I am doing in this problem! Any help is appreciated!

• Let's start with the basics. If [K+] is the concentration of potassium cation and [ClO4-] is the concentration of perchlorate anion, what is the formula for the solubility product, Ksp? – MaxW Apr 29 '20 at 5:57
• Effort for a question can be as little as writing a reaction equation and looking up a few key words. Adding a citation to the exercise also often helps for context. Generally spending some time with your post to make it look nice will go a long way in attracting answers. – Martin - マーチン Apr 29 '20 at 10:24
• @Martin-γγΌγγ³ Thank you for your insight, I will remember that! – DakotaK Apr 29 '20 at 16:25

If we take it generally, e.g. for a reaction $$\ce{A + C <=> C + D}$$

then $$K$$ in $$K = \frac{\ce{[C][D]}}{\ce{[A][B]}}$$ means the reaction equilibrium constant and $$\ce{[X]}$$ is a convention for the molar concentration of $$\ce{X}$$ in mol/L, in this case the equilibrium concentration.

OTOH. very similar quantity $$Q$$ in $$Q = \frac{\ce{[C][D]}}{\ce{[A][B]}}$$ is called the reaction quotient with plugged in the actual concentrations of the involved compounds.

If $$Q \lt K$$, the reaction is ongoing forward, toward the products on the right reaction side.

If $$Q \gt K$$, the reaction is ongoing backward, toward the reagents on the left reaction side.

If $$Q = K$$, the reaction is in equilibrium, with the net reaction rate being zero. It means, on the molecular level, there reaction is ongoing in both directions at the equal rate.

Now, $$K_\mathrm{sp}$$ is a specific equilibrium constant, called a solubility product and $$Q_\mathrm{sp}$$ is a reaction quotient of eventual precipitation/dissolution process.

For the reaction $$\ce{AB(s) <=>[\ce{H2O}]A+(aq) + B-(aq)}$$,

$$K_\mathrm{sp}=\ce{[A+][B-]}$$ is for the equilibrium between solution and precipitate.

$$Q_\mathrm{sp}=\ce{[A+][B-]}$$ is for the current concetration of ions.

If $$Q_\mathrm{sp} \lt K_\mathrm{sp}$$, all precipitate is dissolved and the solution is not saturated yet.

If $$Q_\mathrm{sp} = K_\mathrm{sp}$$, the solution is saturated and the dynamic equilibrium is ongoing between the solution and precipitate.

If $$Q_\mathrm{sp} \gt K_\mathrm{sp}$$, the precipitate is being formed, or the solution is oversaturated, what is ametastable state, that is easily disturbed.

• Thank you for your help, I appreciate your time! That makes sense. I calculated the moles of both K+ and ClO4-, which is .001 moles and .04 moles, respectively. And then, Qsp is simply those two concentrations multiplied together, for which I obtained .00004. Qsp, as a result, is less than the given Ksp so no precipitate forms. That is because Qsp is the value we have obtained from our solution and because it is less than equilibrium, a full chemical reaction will not be able to occur! This helped a lot, it was simpler than I thought it would be, assuming I am on the right track! Thank you! – DakotaK Apr 29 '20 at 16:22
• Thank you for your help, I appreciate your time! That makes sense. I calculated the moles of both K+ and ClO4-, which is .001 moles and .04 moles, respectively. And then, Qsp is simply those two concentrations multiplied together, for which I obtained .00004. Qsp, as a result, is less than the given Ksp so no precipitate forms. That is because Qsp is the value we have obtained from our solution and because it is less than equilibrium, a full chemical reaction will not be able to occur! This helped a lot, it was simpler than I thought it would be, assuming I am on the right track! Thank you! – DakotaK Apr 29 '20 at 16:23