An industry discharges a waste stream that consists of a single monoprotic acid

Question

An industry discharges a waste stream that consists of a single monoprotic acid HA. The titration curve for the titration of the waste stream with NaOH is given below. Unfortunately, the technician forgot to record the HA concentration in the waste stream, the NaOH concentration of the titrant, or amount of base added. The technician is no longer employed you have been hired to figure this out! In the figure below of the titration results, the tick marks are evenly spaced and the x-axis starts at zero (a zero base added). a.) Estimate the pka of HA from the titration curve. b.) What is the total acid concentration (C [HAJ+ CAD in the waste stream? c.) How many liters of l M NaOH are required to neutralize 100,000 L of the waste to 7? 13 12 11 10 Base Added

Explanation / Answer

a). From the titration curve it is clear that equivalence point is reached at 5 unit volume of base added.

pKa of the acid is given by the pH of the acid when volume of base equals the half of the equivalence volume.

Half of the equivalence volume = 2.5 unit volume of base added.

2.5 unit volume of base added corresponds to the pH = 7

Thus pKa = 7.

b). at this point when pH of the solution is 7,

moles of H+ ions = moles of OH- ions.

Thus [HA] = [A-]

Initially pH = 4.5

[H+] = 10-4.5

= 3.162 x 10-5 M

[HA] = 3.162 x 10-5 M

Total acid concentration = [HA] + [A-]

= 2 * [HA]

= 2 * 3.162 x 10-5 M

= 6.324 x 10-5 M

c). Volume of waste water = 100000 L

Moles of H+ = 6.324 x 10-5 M * 100000 L

= 6.324 moles

Moles of NaOH required = 6.324

Molarity of NaOH = 1 M

Volume of NaOH required = 6.324 / 1

= 6.324 L

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