calculating pOH

asked by @kimberlyc39 • about 1 year ago • Chemistry • 5 pts

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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 • about 1 year 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 • about 1 year 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 about 1 year 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 • about 1 year ago