Gennchem q3

[syeh3]

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I know how to do this problem but I don't know how to estimate the cubed root after I divide 1.9 and 4...

 

[emminent]

use the decimals to your advantage. what i mean by that is if it's 10^-13 move your decimal over two #'s.

so for example you have 1.9 in the question. just divide by 4 which i'd round the 1.9 to 2.0 so I'd get .5. now my answer is .5*10^-13.

next i want to make my -13 => -15 so i'd change .5 to 50. my new answer would be 50*10^-15. now just cube root that.

so if i cube 50 then i need to get a good estimate: 3*3*3=27 ; 4*4*4=64 so my answer need be around 3.5 rough estimate

3.5*10^-5 = aprox 4.46

14-4.45=9.54 close enough

 

[shaidester]

Isn't [OH-] = 2x? So once you find molar solubility you do 2*molar solubility to get your [OH-]? Then -log[OH-] = pOH and then 14 - pOH = pH?

 

[emminent]

since this is Ksp the equation should be Ksp = [x][2x]^2 which means ksp = 4x^3

 

[syeh3]

Yep!!! Wow thanks!'n I tott forgot to estimate lol!!

 

[emminent]

No problem; just remember not to forget your stuff on test day. I'd probably be freaking out too but just practice in testing environment. Also once you get at least 3/4 of the work done you should see "Oh it's around 4 so if i subtract from 14 should be around 9."

 

[shaidester]

No problem; just remember not to forget your stuff on test day. I'd probably be freaking out too but just practice in testing environment. Also once you get at least 3/4 of the work done you should see "Oh it's around 4 so if i subtract from 14 should be around 9."

So you don't multiply 2 with the molar solubility (x) to find pOH?

 

[syeh3]

Too kate my each is tomorrow but I won't forget to estimate lol

 

[emminent]

Damn GOODLUCK! Do well and just focus; give us a breakdown on how you did 🙂

So you don't multiply 2 with the molar solubility (x) to find pOH?

What? Ksp = [Mn][OH]^2; there's two OH so Ksp = [Mn][2OH]^2; change it to variables Ksp = [x][2x]^2 which means you square the OH concentration to get Ksp = 4x^3; now plug in values 1.9*10^-13 = 4x^3

 

[Member 139582]

Yea I think Shaidester is right, you multiply the molar solubility (x) by 2, to get the OH- conc. before solving for pOH and pH.

 

[shaidester]

Yeah here is a link: http://www.chemteam.info/Equilibrium/calc-pH-from-Ksp.html

Scroll to problem #2 and look at how this person works out an almost identical problem to this one. Here is the thing they say: "This is an important point: what we have calculated is 'x' and it is NOT the [OH¯]. That value is '2x.'"

 

[emminent]

Ah true; my bad again. Sorry just rushed through these without thinking too much in the morning.