Determine carbonate alkalinity in mg CaCO3/L. Determine total alkalinity in mg C

Question

Determine carbonate alkalinity in mg CaCO3/L.

Determine total alkalinity in mg CaCO3/L.

Estimate bicarbonate (HCO3-) concentration of the water in mg CaCO3/L.

Water was sampled from a cave in Cheju Island. A titration experiment was conducted to determine alkalinity of the cave water. 500 mL water consumed 1 mL of 0.02 N H2SO4 to down pH to 8.0, and further consumed 10 mL of 0.01 N HC1 to down pH to 4.0 Thermodynamic Constants for Species of Importance in Water Chemistry Values of Constants Metric Equivalents of Standard Units

Explanation / Answer

water sample taken=500mL=0.5L

CO32- end point

Volume of 0.02N H2SO4 used=1mL

H2SO4 + CO32-==>HCO3- + HSO4-

(CO32-)mg/L=(0.02*0.001/0.5)*1000=0.04mg/L

CO32- as mg/L CaCO3=0.04mg/L(1/30.0meq/mg)*50mgCaCO3/meq

=0.067mg/L as CaCO3

HCO31- end point

Volume of 0.01N HCl used=10mL

HCl + HCO31-==>H2CO3 + Cl-

(HCO31-)mg/L=(0.01*0.01/0.5)*1000=0.2mg/L

HCO31- as mg/L CaCO3=0.2mg/L(1/61.0meq/mg)*50mgCaCO3/meq

=0.164mg/L as CaCO3

Total alkalinity(mg/L as CaCO3)=[HCO31-] + 2[CO32-] + [OH-] + [H+]

since at pH =6-8 concentration of [H+] and [OH-] is negligible so total alkalinity is due to carbonate and bicarbonate. the alkalinity is the ability of water to absorb H+ ions

Total alkalinity(mg/L as CaCO3)=[HCO31-] + 2[CO32-]=0.164+0.067=0.231mg/L as CaCO3

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