# chemistry

posted by
**Heather**
.

Water in contact with air is acidic due to the dissolved carbon dioxide. Water in equilibrium with the air contains 4.4 x 10^-6% CO2. The resulting carbonic acid, H2CO3,gives the solution a hydrogen ion concentration of 2.0 x 10^-6M, about 20 times larger than that of pure water. Calculate the PH of the solution.

pH = - log [H3O+].

PH = 4.7

But it says 20 times larger than pure water and the PH of water is 7 so the PH of this solution has to be 20 x 7 right?

You must have punched a wrong number because (H^+) of 2.0 x 10^-6 gives a pH of 5.7 and not 4.7. Check to make sure that is correct.

And it IS 20 times larger.

Pure water is 1 x 10^-7 M. This solution is 2 x 10^-6 M.

(2E-6/1E-7)= 20.0 on the nose.

20 x 7 isn't correct because the 7 came from the log term. Change pH 7 to (H^+) first to get (H^+) = 1 x 10^-7, THEN divide into 2 x 10^-6.

so divide 7/5.7 = 1.23

My inclination is to say, "No," but I'll ask why?

The 20 times values comes from

2 x 10^-6/1 x 10^-7 = 20.0.

Dividing 7 by 5.7 isn't correct (if you are trying to get the 20.0) because you are dividing a log value by another log value and log values are not directly comparable. For example, look at the following values.

pH.......(H^+)

3.0.....1 x 10^-3

4.0.....1 x 10^-4

5.0.....1 x 10^-5

6.0.....1 x 10^-6

Notice that changing the pH by just 1.0 units, we change the (H^+) by 10.

but im trying to find PH. Is it 5.7?

yes.

If (H^+) = 2E-6, then

pH = -log(H^+)

pH = -log(2E-6)

pH = -(-5.7)

pH = +5.7

Let me know if you have any problems getting that value.