Can someone check this for me please?

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The discussion centers on the solubility and dissociation of silver sulfate in water, with participants calculating the Ksp value and discussing electrochemical cells. The dissolving equation for silver sulfate is provided, along with the Ksp expression, which is confirmed to be correct. Participants also explore the separation of ions in a solution and the half-reactions involved in an electrochemical cell with nickel and silver electrodes. Additionally, there is a focus on calculating the hydronium ion concentration from acetic acid's dissociation, with discussions on molarity and equilibrium expressions. The conversation reflects a collaborative effort to solve chemistry problems and clarify concepts.
  • #51
okay

Ka = 1.8 x 10^-5= (9.0 x 10^-3)^2
_____________
[CH3COOH]

At = , [CH3COO-] = [H3O+] = 9.0x10-3mol/L

I think...
 
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  • #52
okay that is supposed to be (9.0 x 10^-3)^2 divided by [CH3COOH]
 
  • #53
yoshi6 said:
okay

Ka = 1.8 x 10^-5= (9.0 x 10^-3)^2
_____________
[CH3COOH]

At = , [CH3COO-] = [H3O+] = 9.0x10-3mol/L

I think...
where did you get 9.0 x 10^-3?
 
  • #54
wrong? I don't know I'm very confused, I am trying to follow the examples in my book, but I don't really understand what it going on...
 
  • #55
K_a=\frac{[H_3O^+][CH_3COO^-]}{CH_3COOH}

1.8\times10^{-5}=\frac{x^2}{\frac{[1.2/60.06]}{1.0}}

you wrote your Ka correctly up there, idk what happened.

you're correct that the Acetate ion and Hydronium ion concentrations will be the same.
 
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  • #56
but where does the 60.06 come from? what is that
 
  • #57
yoshi6 said:
but where does the 60.06 come from? what is that
let me fix that, i should have that within another fraction

how do you calculate molarity? what units should/are used for equilibrium concentrations?
 
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  • #58
molarity = moles of solute/ liter of solution
 
  • #59
yoshi6 said:
molarity = moles of solute/ liter of solution
what is the molar mass of Acetic acid and the moles of 1.2 grams of Acetic acid that will be dissolved?
 
  • #60
60.05 g/mol i think
 
  • #61
yoshi6 said:
60.05 g/mol i think
correct, what about molarity and what would you do with it to find the Hydronium concentration at equilibrium?

K_a=\frac{[H_3O^+][CH_3COO^-]}{CH_3COOH}

1.8\times10^{-5}=\frac{x^2}{\frac{[1.2/60.06]}{1.0}}

it could also be this

1.8\times10^{-5}=\frac{x^2}{\frac{[1.2/60.06]-x}{1.0}}
 
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  • #62
oh so that is where you got x^2/ [1.2/60.05]...
 
  • #63
yoshi6 said:
oh so that is where you got x^2/ [1.2/60.05]...
yep ... and have you read when you can neglect x = the concentration change?

i just calculated it and the difference was only by .09
 
  • #64
okay, makes sense, I'm just curious, what time is it over there?
 
  • #65
yoshi6 said:
okay, makes sense, I'm just curious, what time is it over there?
1AM, i'im so bored, lol. what time is it there?

i have class tomorrow but i can't sleep :-[
 
  • #66
wow...2am, I have been doing homework for hours...thank you so much for your help! I feel like I owe you a present
 
  • #67
yoshi6 said:
wow...2am, I have been doing homework for hours...thank you so much for your help! I feel like I owe you a present
naw idc, i want to be a chem tutor @ my school but i don't think i'll get hired b/c they prefer to hire professionals :( so I'm happy to help because i really don't want to forget what i learned, and i have no motivation to review on my own.

you should go to sleep, lol ...
 
  • #68
I should but I don't think I am going to yet...what school do you go to?
 
  • #69
yoshi6 said:
I should but I don't think I am going to yet...what school do you go to?
i go to a CC in Houston, lol ima PM u because i don't think they'd like chatting
 
  • #70
lol okay
 
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