Question

A 25.0-mL sample of lambic (a sour Belgian beer) was analyzed for its acetic acid content by an electrochemical method. The pure sample registered a signal of 2.55 mA. A standard solution of 0.500 mL of 0.109 M acetic acid was added, and the signal increased to 2.98 mA. What is the acetic acid concentration in the original lambic sample?

Answer 1

Ans. Given,

            Signal of pure sample = 2.55 mA

            Signal after spiking with the standard = 2.98 mA

            Increased in signal due to spiking = 2.98 mA - 2.55 mA = 0.43 mA

# Preparation of spike sample: 0.5 mL of 0.109 M standard acetic acid solution is added to 25.0 mL pure sample.

            Final volume of spiked solution = 0.5 mL (std. acetic acid) + 25.0 mL (pure sample)

                                                                        = 25.5 mL

[Acetic acid] in the spiked aliquot can be calculated using -

C1V1 = C2V2

C1= Concentration, and V1= volume of initial solution 1         ; i.e. std. solution,

C2= Concentration, and V2 = Volume of final solution 2         ; Final spiked aliquot

[Acetic acid] in final aliquot, C2 = (C1 V1) / V2

= (0.109 M x 0.5 mL) / 25.5 mL

= 0.002137 M

Therefore, spiking increases the acetic acid concentration in pure sample by 0.002137 M.

# Increase in signal (0.43 mA) of spiked solution must be equivalent to increase in acetic acid concentration by 0.002137 M due to spiking.

Thus,

            0.43 mA signal is equivalent to 0.002137 M

     Or, 2.55 mA signal is equivalent to (0.002137 / 0.43) x 2.55 M

                                                            = 0.0127 M

Therefore, signal of 2.55 mA of pure sample is equivalent to acetic acid concertation of 0.0127 M.

Hence,

            [Acetic acid] in pure sample = 0.147 M