## Use of Graphing Calculator for pH Box Math

[H** ^{+}**] = 2.75 E-6

Start with [H** ^{+}**] = 2.75 E-6.
Input

__(-)__(This is the 'change sign' button.),

__log__(not

*ln*, the natural log)

__2__

__.__

__7__

__5__

__E__

__(-)__

__6__and

__enter__. The pH will be displayed as 5.560667306. It is dreadfully unrealistic to consider that number as a final answer of the pH of the solution because pH is not generally accurately measurable beyond two places to the right of the decimal. If this were a final answer, I would round it to 5.56, but if you want to continue around the pH box, you should keep as many of the digits as you can.

pH = 5.560667306

Punch __(-)__ to change the sign and __ans__ to call up the previous
answer. (This inserts the 5.60667306 into the new calculation.) Punch __+__,
the plus sign and __1__ __4____enter__ to add 14 to the negative pH.
This will give you the pOH of 8.439332694. (Round to 8.44 if this is your final
answer.)

pOH = 8.439332694

To get to the [OH** ^{-}**]
from the pOH, punch in the antilog (usually

__INV__or

__shift__and

__log__), change sign

__(-)__, and the pOH from the previous answer

__ans__

[OH** ^{-}**] = 3.636363636 E-9

To get back to the [H** ^{+}**]
from the [OH

**], enter the Kw, 1 E -14, and divide by the previous answer. Punch**

^{-}__1__

__E__

__(-)__

__1__

__4__

__ÃƒÂ·__

__ans__

__enter__

[H** ^{+}**] = 2.75 E-6

Now for practice, go around the pH box the other way.

The rules are:

- To get pH from [H
^{+}] or to get pOH from [OH^{-}], use the negative log. - To go from [OH
^{-}] to pOH or from [H^{+}] to pH, use antilog of the negative number. - To go from [H
^{+}] to [OH^{-}] or back, first put in the Kw, 1E-14 and divide by the one you are leaving. - To go from pH to pOH or back, subtract the number you have from 14.

Proficiency in pH box calculations requires practice. There is no Wyzant quiz on the pH box because you can make your own exercises, but you will use the calculations in many problems in this acid-base section.