Question

Field application problem – A small manufacturing company wants to locate in Elk Grove. It will hook up to the city water supply, but needs to reduce the tap water pH to 5.5 for a particular production process. A titration test shows that 4 meq/L is needed to do this. The plant manager plans to feed 2.0 N acid into the water line. If the desired flow is 20 gpm, what flowrate of acid is needed? Answer: 0.04 gpm.

Answer 1

Solution:

Here we have to use the formula used to calculate the alkalinity in milliequivalents per liter:

(meq/L) = mLa×N(meq/mLacid) ×1,000 (mL/L)/mLsample

Where

mLa= total volume of the standard acid solution used to reach the equivalence point, in milliliters

N= normality of the standard acid solution used, in milliequivalents per milliliter acid and

mL sample= volume of the sample titrated, in milliliters.

Given alkalinity= 4 meq/L

mLa= total volume of the standard acid solution used to reach the equivalence point, in milliliters

should be calculated from the desired acid flow=20gpm(gallon per minute)

                               1gallon= 3.785L

          Therefore 20gallons= 3.785Lx 20gallons/ 1 gallon= 75.7082L or 757082mL

mLa= 757082mL

N(meq/mLacid)= 2.0N

Substituting the known values we have

(meq/L) = mLa × N (meq/mLacid) ×1,000 (mL/L)/ mLsample

4 (meq/L) = 757082mL ×2(meq/mLacid) ×1,000 (mL/L)/mLsample or

mLsample x 4 (meq/L) = 757082mL ×2(meq/mLacid) ×1,000 (mL/L) or

mLsample = 757082mL ×2(meq/mLacid) ×1,000 (mL/L) / 4 (meq/L)

                     =378541000mL or approximately 400000000mL or 400000 L per minute