10.00 mL of a solution containing a monoprotic acid was placed in a 100-mL volum

Question

10.00 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. 25.00 mL of this diluted acid solution was then titrated with 0.08733 M NaOH, and 19.88 mL of NaOH was required to reach the endpoint. 1. 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.

Explanation / Answer

a) A monoprotic acid means that the acid dissociates to give 1 equivalent of proton and its conjugate base. For instance 1M solution of HCl or acetic acid gives 1M of protons and 1M of chloride ions and acetate ions respectively while 1M of sulphuric acid, which is diprotic will give 2M of protons but only 1M of sulphate ions. Proticity for acids are used to measure the number of ionizable protons in the acid.

b) NaOH is a monobasic base meaning it produces one equivalent of hydroxide ions. Since neutralization involves reaction of hydroxide ions with protons to form water in a 1:1 ratio, neutralization of monoprotic acid by a monobasic base like NaOH will require 1 equivalent of NaOH. This means that the no.of moles of NaOH consumed is equal to the no.of moles of acid present here. Thus the molar ratio of acid to NaOH is 1:1.

c) 0.08733M of NaOH contains 0.08733moles of NaOH in 1L of solution or 1000mL. Thus using unitary method, the moles of base in 19.88mL is (0.08733/1000)*19.88 = 1.7361mmoles of NaOH (1000mmol = 1mol).

d) Using the law of equivalence, the volume and molarity of the base consumed is used to calculate the molarity of acid as VaMa = VbMb => Ma = (VbMb)/Va where a and b subscripts stand for acid and base. Now upon substituting the known values, we get molarity of acid in the diluted solution which is titrated as (19.88*0.08733)/25 = 0.069445M.

e) Since the original acid solution was diluted from 10mL to 100mL, the original solution is 10 times more concentrated than the one taken for titration. As molarity does not vary with volume of the same solution, the molarity in 25mL of the acid is what is present in the 100mL diluted solution also. Thus the original will be 10 times greater than that which gives the concentration of the original solution as 0.69445M.

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