Can someone check this for me please?

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SUMMARY

The discussion revolves around the solubility product constant (Ksp) of silver sulfate (Ag2SO4) and its calculations. The solubility of silver sulfate is established at 1.3 x 10^-6 mol/L at 20 degrees Celsius. The Ksp expression is defined as Ksp = [Ag+]^2[SO4^2-], leading to a calculated Ksp value of 1.7 x 10^-12. Participants also explore related concepts such as electrochemical cells, ion separation procedures, and the calculation of hydronium ion concentration in acetic acid solutions.

PREREQUISITES
  • Understanding of solubility product constants (Ksp)
  • Knowledge of electrochemical cells and half-reactions
  • Familiarity with acid-base equilibrium and Ka expressions
  • Basic concepts of ion concentration and molarity
NEXT STEPS
  • Study the calculation of Ksp for various salts, focusing on silver sulfate.
  • Learn about electrochemical cells, including anode and cathode reactions.
  • Explore the concept of acid dissociation constants (Ka) and their applications.
  • Investigate ion separation techniques based on solubility rules and activity series.
USEFUL FOR

Chemistry students, educators, and anyone involved in analytical chemistry or chemical education will benefit from this discussion, particularly those focusing on solubility and electrochemistry.

yoshi6
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Can someone check this for me please??

The solubility of silver sulphate is 1.3 x 10^-6 mol/L at 20 degrees celsius.

a) write the equation for the dissolving of this salt in water.
b) show the Ksp expression.
c) calculate the value of the Ksp at 20 degrees celsius.

ANSWERS

a) 2Ag+(aq) + SO4 2-(aq) = Ag2(SO4)

b) Ksp = [Ag+]^2[SO4^2-]

c) I am having trouble withthis: Ksp= (1.3 x 10^-6)(1.3 x 10^-6) = 1.7 x 10^-12
 
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lol wrong place. let me work it out.
 
lol thanks...and I'm a girl! :)
 
yoshi6 said:
lol thanks...and I'm a girl! :)
lol i thought about that when i said bro, so i edited my comment but you checked it fasttt! haha

anyways, so I'm assuming C is your work?

Ag_2SO_4 (s) \rightarrow 2Ag^{+} (aq) + SO_4^{2-} (aq)

You're given K_{sp}=1.3x10^{-6} M of Silver Sulfate, and so when you write your K_{sp} equation it will look like this. (You have it correct, I'm just blabbering)

K_{sp}=[Ag^{+}]^2[SO_4^{2-}] and Ag_2SO_4 (s) is not included b/c we don't include liquids/solids since their concentration change is negligible.

Let x=1.3x10^{-6}, x=[2S]^2

simplifying and skipping steps

1.3x10^{-6}=4S^{3}

S=\sqrt[3]{\frac{1.3x10^{-6}}{4}}

and that's all.
 
Last edited:
Ok. Well first, I have this one question I have posted before. I really have no clue what they are asking me, I do not know what to do or where to start. Maybe if you have any ideas I can work with it...

A solution contains the following ions, each at a 0.1 M concentration.

Ag+ (aq), Ca2+ (aq), Ni2+ (aq)

write out a procedure by which these ions may be separated from each other and from the solution, clearly indicating the order of separation and when filtering should be done.

So, what is this procedure and order of separation they are talking about? I have no clue what is going on here.
 
i'm not very good at these type of questions (i didn't care for labs too much, I'm lazy), but what comes to mind is solubility rules & activity series of metals.

i'm sure someone will respond :-]
 
OK, and I am having the same kind of issues with this one:

An electrochemical cell is constructed by placing a nickel electrode into a 1.0 M NiSO4 solution, and a silver electrode into a 1.0 M AgNO3 solution, and then joinging them with a salt bridge to complete the circuit.

a) give the anode half reaction and the E0 value.
b) give the cathode half reaction and the E0 value
c) give the net reaction and its E0 value
 
i think i have an idea

Cl^{-} is soluble except in the presence of Ag^{+} in which it is a solid.

so add Cl^{-} to react with Ag, so now you've taken care of Ag -> AgCl (s)
 
yoshi6 said:
OK, and I am having the same kind of issues with this one:

An electrochemical cell is constructed by placing a nickel electrode into a 1.0 M NiSO4 solution, and a silver electrode into a 1.0 M AgNO3 solution, and then joinging them with a salt bridge to complete the circuit.

a) give the anode half reaction and the E0 value.
b) give the cathode half reaction and the E0 value
c) give the net reaction and its E0 value
read this thread real fast https://www.physicsforums.com/showthread.php?t=181411

and i will work on your problems while thinking of some questions to ask
 
  • #10
ok i got it worked out, my question is. what is the purpose of Sulfate and Nitrate?

what are the charges for the individual ions? Ni, SO4, Ag, NO3?
 
  • #11
The common charges?

Ni2+, Ag1+, SO4 2-, NO3 1-...right?
 
  • #12
yoshi6 said:
The common charges?

Ni2+, Ag1+, SO4 2-, NO3 1-...right?
EDIT: lol ... you have it right, idk what I'm thinking, anyways moving on

so did you read the other thread?

so now look at the reduction potential table and tell me which one you would choose to be your cathode and anode?
 
Last edited:
  • #13
lol btw the four of them?
 
  • #14
or just between NiSO4 and AgNO3
 
  • #15
I would have to say AgNO3 cathode...
 
  • #16
you didn't answer my question, what is the purpose of Nitrate and Sulfate?

look at your reduction potential table, are you able to find them?

why do you choose AgNO3 as your cathode?
 
  • #17
I am so confused
 
  • #18
yoshi6 said:
I am so confused
sigh :-[ ok let's move on

a rule of thumb, E_{cell}>0, we need to choose which one will be the cathode/anode so that E_{cell} is positive.

tell me the what the half-reactions for Ag^{+} and Ni^{2+}.
 
  • #19
Okay, like so?

Ag -> 2g-
is that right for Ag+
 
  • #20
Ag-> Ag+ + e- is oxidation right?
 
  • #21
yoshi6 said:
Okay, like so?

Ag -> 2g-
is that right for Ag+
ok from this list, which half-reactions for Ag+ and Ni2+ would you choose?

tool_g02.gif
 
  • #22
yoshi6 said:
Ag-> Ag+ + e- is oxidation right?
yes, correct.
 
  • #23
ok so our half-reactions are

Ag^+ (aq) + e^- \rightarrow Ag (s) \ E^0 +0.80

Ni^{2+} (aq) + 2e^- \rightarrow Ni (s) \ E^0 -0.28

correct?
 
  • #24
Okay so
Ni2+(aq) + 2e- ---> Ni(s) - .28

Ag+(aq) + e- ------> Ag(s) .799
 
  • #25
lol wow, I'm embarressed
 
  • #26
yoshi6 said:
Okay so
Ni2+(aq) + 2e- ---> Ni(s) - .28

Ag+(aq) + e- ------> Ag(s) .799
now balance your equation so that the amount of electrons being transferred is the same.

then choose which will be the anode b/c they are currently both in their reduction state.

E_{cell} > 0
 
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  • #27
yoshi6 said:
lol wow, I'm embarressed
don't be :-] i don't know anything about labs so I'm useless there :-D
 
  • #28
sorry my computer keeps shutting down...one minute
 
  • #29
btw, i think most tables are ordered from the most positive value (reduction, oxidizing agent) to the most negative (oxidized, reducing agent)

so be careful
 
  • #30
okay is this one right?
2Ag + 2e- -> 2Ag0

isnt the cathode reaction: Ag+ + e- -> Ag
and the anode reaction: Ag-> Ag+ + e-

Is that significant at all?
 

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