What is the recommended way of preparing 100ml of 0.02m from 2m naoh in the laboratory?

The recommended way of preparing 100ml of 0.02 M from 2M NaOH in the laboratory is to obey the equation V1S1=V2S2.

Explanation:

According to the given question

  V2=100ml

  S2=0.02M

   S1= 2M

 Then V1=?

     We all know that V1S1=V2S2

                                 or  V1=V2×S2÷S1

                                  or  V1= 100×0.02÷2

                                  or  V1 =1  ml

It can be concluded that 1ml of 2M NaOH solution and 99 ml distilled water should be mixed to prepare 100ml of 0.02 M NaOH solution.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.