Question

Would precipitation occur when 500 mL of a 0.02M solution of \(AgNO_3\) is mixed with 500 mL of a 0.001M solution of NaCl? Show your work.

Answer 1

@sammixboo can you come help?

Answer 2

@EclipsedStar

Answer 3

@JFraser

Answer 4

Can you show me what to do?

Answer 5

To figure out if a precipitate will form, we need to see if the Q for this reaction is greater than or less than the Ksp. In order to set up the right equation, we first need to to set up the right chemical reaction. What is the balanced reaction for AgNO3 with NaCl?

Answer 6

what's Q and Ksp? sorry, I'm really bad at this.

Answer 7

well I don't have a periodic table with me but so far I have AgNO3+NaCl---> AgCl+NaNO3

Answer 8

I'm not sure if the products are correct.

Answer 9

@matt101

Answer 10

are you still going to help me?

Answer 11

Yup sorry just got called away for a minute

Answer 12

okay.

Answer 13

In general K is the equilibrium constant (given by product equilibrium concentrations over reactant equilibrium concentrations). Depending on the reaction, a subscript is tacked on to describe that reaction. For this sort of reaction, we use Ksp, where sp refers to the solubility product (because this is a "dissolving" reaction).

Q is calculated exactly the same way as K, but using whatever concentrations you have (not necessarily equilibrium concentrations). If Q > K, the reverse reaction will be favoured until K is reached. If Q < K, the forward reaction will be favoured until K is reached.

Your equation above is correct. Which product is the precipitate?

Answer 14

I don't know...

Answer 15

In this case it's AgCl.

You should know that NaNO3 dissolves easily from your solubility rules - Group I compounds (Na is Group I) always dissolve, as does NO3- containing compounds. That means NaNO3 won't precipitate here, leaving AgCl.

Since AgCl is the precipitate, this is the compound we focus on for this question (Na+ and NO3- are just spectator ions). We want the "dissolving" reaction for AgCl, which is:

\[AgCl \rightarrow Ag^+ +Cl^-\]
For the concentrations of Ag+ and Cl- that we have, we need to figure out which way the reaction will proceed (by calculating Q and comparing it to K). If the reverse reaction is favoured, AgCl will form and precipitation will occur. If the forward reaction is favoured, Ag+ and Cl- will remain dissolved and no precipitate will form.

So far so good?

Answer 16

I think so... so how do I fink K and Q?

Answer 17

@matt101

Answer 18

what do you mean by reverse reaction and forward reaction?

Answer 19

Forward: AgCl --> Ag+ + Cl-
Reverse: Ag+ + Cl- --> AgCl

Answer 20

We need to find the concentrations of Ag+ and Cl- we're working with, then plug that into the equation to calculate Q.

We have 500 mL of each solution, but if we mix them we'll have 1000 mL or 1 L total. This means the concentrations will change. You can figure out these new concentrations using M1V1=M2V2.

Answer 21

ummm I'm sorry, but how?

Answer 22

@matt101

Answer 23

M=concentration, V=volume

Your starting concentration times your starting volume equals your final concentration times your final volume.

For Ag+, the starting concentration is 0.02 M, and the starting volume is 0.5 L. The final volume is 1 L. Use these values to solve for final concentration.

Then repeat the process for Cl-, but using the values for Cl- instead!

Answer 24

ohhhh... .02*.5=1*X
so the final concentration is 0.01?

Answer 25

@matt101

Answer 26

Right! Now do the same for Cl- using the appropriate values.

Answer 27

is it 0.0001*.5?

Answer 28

oops 0.001*0.5?

Answer 29

I got 1/2000

Answer 30

That's right! Let's call it 0.0005 M for Cl-
And we had 0.01 M Ag+

Now the Ksp equation for our equation is:

\[K_{sp}=[Ag^+][Cl^-]\]
AgCl does not appear in the equation because it is a pure substance (a solid). What do you get when you multiply the concentrations together?

Answer 31

do I multiply 0.01 and 0.0005?

Answer 32

That's right

Answer 33

1/200000

Answer 34

right?

Answer 35

now what do I do?

Answer 36

Compare this value to your Ksp. You don't have a Ksp in your question, but you must have it somewhere in your notes or maybe in a previous question/table?

Answer 37

nope... that's just the question... it was only one question.

Answer 38

Hmm well the only way to answer this question is if we have the Ksp.

I just looked it up online and the Ksp for AgCl is 1.8 x 10^(-10).

Is your value for Q that you just calculated bigger or smaller than this number?

Answer 39

bigger? because it's 5*10^-6

Answer 40

is that right?

Answer 41

Yes! So if Q > Ksp, which way will the reaction proceed?

Answer 42

umm forward?

Answer 43

@matt101

Answer 44

Nope - see what I wrote above.

Keep in mind that Q and Ksp are calculated by products/reactants. If Q is bigger than Ksp, we need to decrease products and increase reactants to make Q smaller and reach equilibrium - this means the reverse reaction will be favoured, since it uses up products and forms reactants.

The exact opposite would occur if Q is less than Ksp.

Does that make more sense now?

Answer 45

ummm not really. sorry

Answer 46

No problem - what are you having trouble understanding?

Answer 47

"Keep in mind that Q and Ksp are calculated by products/reactants. If Q is bigger than Ksp, we need to decrease products and increase reactants to make Q smaller and reach equilibrium"

Answer 48

Here's the equation for Ksp:

\[K_{sp}={[P]_{eq} \over [R]_{eq}}\]
Using the EQUILIBRIUM concentrations of products (P) and reactants (R).

Q is calculated in the same way, but using whatever concentrations are given (since we don't know if the system is at equilibrium):

\[Q={[P] \over [R]}\]
If the value we get for Q is BIGGER than Ksp, it must be because the numerator is too big (too many products) and the denominator is too small (too few reactants). Since the system must move towards equilibrium, the only way for this to happen, for Q to equal Ksp, is to decrease the amount of products and increase the amount of reactants. This means the reaction will move in the reverse direction so that this occurs. If you start calculating Q every so often after the reaction begins, you'll find its values gets lower and lower until Q equals Ksp, and at this point the system is at equilibrium.

If the value we get for Q is SMALLER than Ksp, the exact opposite occurs (and the forward reaction will be favoured).

Does it make more sense now? Read it over carefully a few times to make sure you get what I'm saying here.

Answer 49

ohhhhh.. so how do I know if precipitation occurs?

Answer 50

@matt101

Answer 51

The precipitate is the reactant. The products are dissolved (aqueous).

If the reaction is going towards the reactants, precipitation occurs.
If the reaction is going towards the products, precipitation does not occur.

You said Q > Ksp. Which way is the reaction going if this is the case?

Answer 52

reverse

Answer 53

right?

Answer 54

@matt101

Answer 55

That's right! So about the precipitate...?

Answer 56

Yes? because Q is bigger than K?

Answer 57

You got it!

Answer 58

okay, thank you