# chem

posted by
**Jon** on
.

this is my question

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

I don't 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 10^{x} 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^-)**