Friday
July 25, 2014

Homework Help: chem

Posted by Jon on Friday, March 23, 2007 at 9:40pm.

this is my question
Calculate the hydronium ion consentration and the hydroxide concentration in blood in which the PH is 7.3.

I dont have invlog on my calculator so I hit log(7.3)= .863 for hydronium. Is this dumb for me to do?

That is dumb for you to do.

Get a calc that has an INV key. Mine costs eight dollars.

concentration= invlog (-7.3)= 5.01?E-08

Do you have a 2nd key, and a log key?

normally, 2nd Log is same as INVLOG

yes when I press 2nd then log I get 10^, is this correct?

In order to get antilog on mine I punch the 10x key. And a pH of 7.3 is hydroniuim of 5.01E-8.

ok I got that part. now how can I find Hydroxide concentration from this?

Two ways.
#1. (H^+)(OH^-) = 1 x 10^-14.
You know (H^+).

#2. pH + pOH = pKw = 14
You know pH, so subtract from 14 to get pOH. Then pOH = -log(OH^-).



PH + pOH = pKw =14
7.3 + pOH =pKw =14
7.3 - 14 = 6.7
pOH = -log(OH^-)
6.7 = -log(1.00 x 10^-14)
pOH = 1.92 x 10^6? I'm stuck


PH + pOH = pKw =14
7.3 + pOH =pKw =14
7.3 - 14 = 6.7
pOH = -log(OH^-) OK to here

6.7 = -log(1.00 x 10^-14)
pOH = 1.92 x 10^6? I'm stuck


pOH = -log(OH^-)
6.70 = -log(OH^-)
-6.70 = log(OH^-)
1.996 x 10^-7 = round to
2 x 10^-7 = (OH^-)

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