Gennchem q3
Started by syeh3
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
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
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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?
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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?
Damn GOODLUCK! Do well and just focus; give us a breakdown on how you did 🙂
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
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
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.
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.'"
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.'"
