Question

20 ml of 0.2 m naoh is added to 50 ml of 0.2 m ch3cooh hence ph-pka

Answer 1

In calculation of pH of a given reaction, first you need to write the equation for the reaction down then determine the equilibrium concentrations after which you can go ahead by calculating the pH.

So for this case we have been give the volume of sodium hydroxide as 20 ml and its concentration as 0.2 molar.

On the other hand, concentration of ethanoic acid is 0.2 molar and its volume is 50 ml.

Then calculate the number of moles in each case.

Moles of NaOH = 0.2 x 20/1000 = 0.004 moles

moles of CH₃COOH =0.2 x 50/1000 = 0.01

so ethanoic acid is in excess of (0.01 - 0.004) = 0.006 moles

the total volume = 20 + 50 = 70 ml

then concentration = 0.006/70

                               = 0.0000857

                   then pH= -log 0.0000857

                            pH= 4.1

but ka of ethanoic acid is 1.8 x 10⁻⁵

            Hence pka = 4.7

Therefore we can conclude that pH is less than pka.