Question

After the second addition of 6M NaOH to the test tube, what is the color of the solution (Pink) ? IS this the color you expected the solution to be? Explain your reasoining behind this answer and include equations for reactions in your explanation.

Answer 1

For strong base like NaOH (6M) , the indicator used is phenolpthalein which gives pink color under basic conditions.When in a titration ,base is slightly in excess than the acid, the acid is neutralized completely and the excess base (with phenolpthalein) gives pink color to the solution.

Phenolpthalein indicator works in the pH range 8.2-12.In strongly basic conditions ,again phenolpthalein turns colorless/(for pH>13)

pH=-log [H+]

In 6M NaOH ,NaOH completely dissociates to give Na+ and OH- ions

[NaOH]=[Na+]=[OH-]=6M

Also [H+][OH-]=10^-14 (ionic product of water)

So.[H+]=10^-14/[OH-]=10^-14/6=1.7*10^-15

pH=-log (1.7*10^-15)=14.8

pH=14.8

So,the color of the solution should be colorless under strongly basic conditions pH>13