Answer to Question #303810 in Physical Chemistry for Chemistry

Question

A Laboratory assistant prepares 5 litres of 0.1N hydrochloric acid stock solution using commercial HCl available in laboratory which has percentage purity of 35% (w/w) and specific gravity of 1.18 g/ml. Then he fills five 100ml standard flasks with 50.0ml of 0.1N HCl solution and dilutes it with distilled water upto the mark. Then he takes 10ml of diluted HCl solution from each flask into a conical flask, mixes it and adds 50ml of 0.025M Sodium hydroxide solution to it. Finally, he adds few drops of phenolphthalein to the resulting solution in the conical flask.

i) What was the concentration of HCl solution prepared in 100ml standard

flasks?

ii) What do you think will be the colour of the resulting solution after addition of

the indicator?

iii) Calculate the pH of the resulting solution.

Answer 1

i)                   What was the concentration of HCl solution prepared in 100ml standard flasks?

Standard stock solution concentration: 0.1

Volume taken: 50

Final volume: 100

C1V1 = C2V2

C2 = C1V1/V2 = 0.1*50/100 = 0.05 (M)

 

ii)                  What do you think will be the colour of the resulting solution after addition of the indicator?

Moles HCl: 10 mL (0.01 L) of 0.05 M is taken, so 0.01 * 0.05 = 0.0005 mol

Moles NaOH: 50 mL (0.05 L) of 0.025 M is taken, so 0.025 * 0.05 = 0.00125 mol

HCl + NaOH = NaCl + H₂O

NaOH is 2.5 times in excess, so the medium is basic (pink colour)

 

iii)                Calculate the pH of the resulting solution.

0.005 mol of HCl and NaOH is used up

0.00125 – 0.0005 = 0.00075 mol NaOH left

pOH = -log₁₀ (0.00075) = 3.12

pH = 14 – pOH = 10.88