Which of the following choices is NOT a potential application of the solubility product constant?
A. Use (K sp) to find solubility.
B. Determine ion concentrations that will prevent a precipitate from forming.
C. Determine reactant concentrations that will allow a precipitate to form.
D. Determine ion concentrations that will allow a precipitate to form.
I'm not sure about this problem. Can someone please help me out?
The answer was actually C.
He actually knows what he's talking abt, unlike us students who need jiskha to steal answers from
Well if he knew what he was talking about then he would actually given the correct answer, and plus the logic of the that answer doesn't even make sense DRBOB.
STOP USING RANDOM INLOGICAL FACTS AND FOCUS ON THE QUESTION, otherwise we dont need ur help
To determine which of the choices is NOT a potential application of the solubility product constant (Ksp), let's examine each option.
Option A states that we can use the solubility product constant (Ksp) to find solubility. This is a valid application of Ksp because the solubility product constant relates to the extent to which a compound dissolves in a solution.
Option B suggests using the solubility product constant (Ksp) to determine ion concentrations that will prevent a precipitate from forming. This is also a valid application because a high concentration of ions can prevent precipitation by exceeding the solubility product constant.
Option C mentions determining reactant concentrations that will allow a precipitate to form. This is another valid application of Ksp because a precipitate will form when the ion concentrations exceed the solubility product constant.
Option D states that we can determine ion concentrations that will allow a precipitate to form. This is essentially the same as option C and implies a valid application of the solubility product constant.
Therefore, the correct answer is none of the above. All of the given choices are potential applications of the solubility product constant.
Thanks but I don't buy it. I can use trial and error and make C work.
Say Ag^+ is 0.1 M, and Cl^- is 1.7E-9 M so Qsp = 0.1 x 1.7E-9 = 1.7E-10 which means the solution is saturated with AgCl. Anything over either of those amounts will form a ppt.
The problem with this problem is that the word potential is used and that allows quite a bit of leeway. I would choose D.
For a salt such as say BaSO4, then
....................AgCl ==> Ag^+ + Cl^-
and Ksp = (Ag^+)(Cl^-)
A. We know that we can determine solubility of AgCl with Ksp = (x)(x) answer is not A.
C. We can compare Qsp with Ksp (< = >) and determine if a ppt will form so I would rule out C.
D. We can use Qsp for this too so I rule out D.
B seems to be shady to me so I would go with that; however, with the word potential I can make B work too. So to me there is no GOOD answer. How would I make B work. I potentially could choose a value for Ag^+, then use Ksp to solve for Cl^- that would just cause pptn, then 1 ion less than that will not give a ppt. If I were grading this test I would not allow it on an exam. That word potential messes up everything because potentially we can do almost anything. Sorry I got on my soap box.