pH, pOH, log???

[CarFanatic]

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Hey guys,
I need a little help on this topic of General Chemistry. I cant seem to grasp the questions that deal with problems that say, find the pH or pOH by giving you like pKa or pKb? And also, I dont understand the whole "log" part and the Henderson Hasselbach equation. Can someone help me by providing a little input and clarification on these topics?

Thank you

 

[yalla22]

i'm so confused about this stuff too🙁

I was hoping there may be a website online that would clear this up but i haven't found any.

 

[71263]

Im not sure exactly what your question is, but I can give you a general overview.

pH- log (concentration of H+ ions) in any given a solution. This generally goes from 1-14. 1-6.9999 is acidic, 7.000001-14 is basic, while 7 is neutral.

pOH- 1-pH, indicates basicity in the same way that pH indicates acidity

pKa, the log of the acid constant, Ka (= [A-][H+]/[HA]), the larger this number is, the stronger the acid, as more H+ ions will have dissociated. Kb, the base constant, is the base equivalent of the Ka.

The H-H equation is only used for solutions of weak acids, where there is only a degree of dissociation. The equation itself is: pH = pKa + log (/[A]). The plus sign is there because if there is more base, it would make sense for the pH to rise. If there is more acid, the log is negative and the resulting pH is lower.

Hope that clears things up!

 

[sammyhagar19]

When converting from a concentration to a pX value you use the -log[x] equation right, so this is the way i think about it

If you have 1 * 10^-1 amount of [H+] the pH is 1 or just the value of the exponent.

Similarly 1 * 10^-5 amount of [H+] the pH is 5

Henderson-Hasselbach is just used to help you figure out how much the pH changes in a buffer solution when you add strong acid or base. (Buffer solutions are just solutions rich in both a weak acid or base and the conjugate of that acid or base which comes from a salt. The weak acid/base and the conjugate can absorb large numbers of H+ or OH- ions put in the solution and thus the pH does not change much). So like above to find the pH or pOH of a buffer solution just find the pKa or pKb by looking at the exponent (if main number i.e. not the exponent is larger than 1, then the pKa will be between the value of the exponent and the next whole number below it) and add the exponent of the [conjugate base]/[weak acid] these concentrations after the strong acid has been added.

 

[vmp200]

Strong Acids: HX ---------> H+ + X-
therefore u can use the pKa to find [H+]:
pKa = [H+][X-]/[HA]
[HA] is usually given. lets take 1M HCl for example with pKa of 4 x 10^-3.

-> 4x10^-4 = [x][x]/[1 Molar HCl]. We use [x] instead of [H+] and [Cl-] because they are formed in equal amounts.

-> sqrt((4x10^-4)*1) = [x]
-> 2x10^-2 = [x]
-> pH = -log[x] = 1.8

Sorry hope this helps, i didnt fully read other replies.