Question

5.00 mL of a solution containing a monoprotic acid was placed in a 100-mL volumetric flask, diluted to the mark with deionized water and mixed well. Then, 25.00 mL of this diluted acid solution was titrated with 0.08765 M NaOH. 9.23 mL of NaOH was required to reach the endpoint. a. What does the term monoprotic mean? b. Determine the molar ratio between the acid and NaOH. c. Calculate the moles of NaOH used in this titration. d. Calculate the molarity of acid in the diluted solution before titration. e. Calculate the molarity of acid in the original solution.

Answer 1

a) Monoprotic acid  

Acid which can donate only one proton to base is called mono protic acid.

Example : HCl is a mono protic acid , it donates only one proton on reaction with base.

Consider two acids H₃PO₃ it contain three hydrogen atoms but it donates only two protons to base so it is a diprotic acid.

H₃PO₄ contain three hydrogen atoms and it donates three protons to base so it is a triprotic acid.

b) Consider a reaction of mono protic acid with NaOH

HA + NaOH $\rightarrow$ NaA + H₂O

From reaction , it is clear that one mole of acid HA reacts with one mole of NaOH.

SO molar ratio of HA: NaOH = 1:1

c) We know that, Molarity = no of moles / volume of solution in litre.

Therefore, no of moles = Molarity x volume of solution in L

No of moles of NaOH = 0.08765 mol / L x 0.00923 L

= 0.000809 mol

No of moles of NaOH used in titration = 0.000809 mol  

The molar ratio of HA : NaOH = 1:1

Hence , no of moles of HA used in titration = 0.000809 mol

25 ml diluted solution have 0.000809 mol HA .

Therefore, 100 ml diluted solution have ( 100 ml x 0.000809 mol HA / 25 ml = 0.00324 mol HA)

5 ml solution diluted to 100 ml in procedure of titration , so moles of HA present in 100 ml diluted solution must be equal to moles of HA present in 5 ml diluted solution.

Therefore , Molarity of 5 ml solution = no of moles / volume of solution in L

= 0.00324 mol / 0.005 L

= 1.62  x 10 ^-05 M