[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 4enter 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 ÷ 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 Chemtutor Quickquiz 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.




Numbers and Math
Units and Measures
Atomic Structure
Periodic Table
States of Matter
Mols, Stoichiometry, and Percents
Oxidation and Reduction Reactions
Solutions beginning
Solutions - Other types of mixture
Acids and bases(top)
Acids and bases, pH box

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