Question

-In the early seasons of Breaking Bad, Walt and Jessie can frequently be seen testing the pH of their downstream product using commercially available indicators intended for pool water testing. When might it be advantageous to use an indicator vs a pH electrode?

-A student tested a solution with a pH meter and recorded the pH to be 7.43. Calculate the [H3O+] and [OH-] for this solution. (Show your work!)

Answer 1

A pH meter is a device that measures the acidity or alkalinity of a liquid.

pH indicators are substances, usually a weak acid or base, that changes color depending on the pH of the solution it is mixed with. Indicators don't change color sharply at one particular pH level. They change over a narrow range of pH. And this color change interval is different for each indicator. For optimal accuracy, the narrower the pH range of the color change the better.

Advantages of using pH indicators

• Liquid indicators are mainly used for titration.
• Easy to use.
• Cheap compared to a pH meter and electrode.
• They are well suited for measurements in drinking water, boiler water, wastewater, surface water, aquarium water, swimming-pool water, and soil samples.

Disadvantages of using pH indicators

• The measured solution should be colorless and clear.
• The accuracy is lower compared to the use of a pH electrode.

Advantages of Using a pH Meter

• Reading a pH meter is much less subjective than reading a color strip or a pH indicator.
• pH meters are much more precise than other methods of pH measurement and are able to measure down to the 0.01th of pH unit.
• pH meters offer the advantage of not being disposable.
• They measure more accurately than a test strip or other method of pH measurement.
  
  
  

The pH is defined as the negative logarithm of [H₃O⁺] or -log[H₃O⁺].

pH = –log[H₃O⁺]

or, [H₃O⁺] = 10 ^-pH

Now, pH = 7.43

so, [H₃O⁺] = 10 ^-7.43

       [H₃O⁺] = 3.715X 10^-8

Now, pH + pOH = 14

           7.43 + pOH = 14

           pOH = 14 – 7.43

           pOH = 6.57

pOH = - log[OH^-]

or, [OH^-] = 10 ^–pOH

     [OH^-] = 10 ^-6.57

       [OH^-] = 2.69 X 10^-7

Or, [H₃O⁺] * [OH^-] = 1 X 10^-14

     (3.715X 10^-8) * [OH^-] = 1 X 10^-14

     [OH^-]= (1 X 10^-14) / (3.715X 10^-8)

     [OH^-]=0.269 X 10^-6 = 2.69 X 10 ^-7

So, [H₃O⁺] = 3.715X 10^-8 and [OH^-] = 2.69 X 10^-7