Question

What is molarity of a solution of NaOH prepared by diluting 20.00 mL of 6.0 M NaOH to 500.0 mL?

Why can we not calculate the actual molarity of the NaOH solution as indicated in the problems above?

Why is an indicator required in the reaction vessel? Which indicator is used?

Answer 1

We have a formula M1V1 = M2V2. Where M1 and M2 are molarities and V1 and V2 are volumes.

Given M1= 6.0 M V1 = 200 ml , M2 =? V2 = 509ml

M2 = 6.0*200/500 = 2.4 M.

The actual molariy of NaOH is 2.4 M.

Sodium hydroxide is a secondary standard. It is highly hygroscopic in nature and absorbs moisture from the surrounding. Hence it's concentration keeps on changing with time. This is why sodium hydroxide solution is standardized using some primary standard.

Usually acid base neutralization reaction is colourless and we will not be able to detect the endpoint of the reaction with our eye. Hence, indicators like phenolphthalein is added to note the endpoint of the reaction.