calculating pOH

asked by @kimberlyc39 • 10 months ago • Chemistry • 5 pts

enter image description here

Add comment
3 answers

14=pH+pOH. For example, for solution a, you will take: 14=12.14+pOH. pOH=1.86.

In order to get number of [OH-] from pOH, take the given pOH value and take 10^-#. For example, in solution d, take 10^-4.13 to get your [OH-]. The relationship between [H+] and [OH-] is that Kw=[H+][OH-]. Kw=1.0x10^-14.

answered by @jennyl31 • 10 months ago

thanks i got it but it keeps saying that [OH] for solution C is not 2x10^-9 and i also tried 2.0x10^-9. what is the answer then ? what am i doing wrong ?

answered by @kimberlyc39 • 10 months ago
  • maybe it's just due to rounding off, try not to round off too much before you arrive at the final answer. Leo commented 10 months ago

Hey there! You seem to be on the right track for the first few questions. I'll just answer C. D will be pretty much the same.

So basically you need these equations:

pH+pOH = 14 (eqn1)

[H+][OH-] = 1x10^-14 (eqn2)

pH = -log(H+)(eqn3)

pOH = -log(OH-) (eqn4)

For C you start of with pH so you can easily get pOHfrom eqn1:

5.38+pOH = 14

pOH = 14-5.38 = 8.62

Now since you have pH and pOH just use eqn 3 and 4:

pH = -log(H+)

5.38 = -log(H+)

[H+] = 10^(-5.38)=4.169x10^-6

pOH = -log(OH-) (eqn4)

8.62 = -log(OH-)

[OH-] = 10^(-8.62)=2.399x10^-9

I hope this helps!

answered by @leot1 • 10 months ago