Question #b7836

1 Answer
Apr 30, 2015

The phenolphthalein alkalinity is #"15.1 meq CaCO"_3"/L"#; the total alkalinity is #"23.6 meq CaCO"_3"/L"#.

Alkalinity in water is caused by the presence of #"HCO"_3^-#, #"CO"_3^(2-)# and #"OH"^-# ions.

Phenolphthalein alkalinity is determined by titration to a phenolphthalein end point (pH 8.3).

It indicates the total hydroxide and half the carbonate present.

Titration to a methyl orange end point (pH 4.3) measures total alkalinity.

The total alkalinity includes all alkalinity due to #"CO"_3^(2-)#, #"HCO"_3^-# and #"OH"^-#.

Alkalinity is usually measured in units of #"mg CaCO"_3"/L"# or #"meq CaCO"_3"/L"#.

The important concept when working with equivalents is that

#"1 eq of anything" = "1 eq of anything else"#.

The alkalinity kits usually give you a formula to use, but let's do the calculations as a chemist would.

Phenolphthalein Alkalinity

Step 1. Calculate the equivalents of #"CaCO"_3#.

#7.55 cancel("mL HCl") × (0.020 cancel("mmol HCl"))/(1 cancel("mL HCl")) × (1 cancel("meq HCl"))/(1 cancel("mmol HCl")) × ("1 meq CaCO"_3)/(1 cancel("meq HCl")) = "0.151 meq CaCO"_3#

Step 2. Calculate the phenolphthalein alkalinity

#("0.151 meq CaCO"_3)/(10.0 cancel("mL")) × (1000 cancel("mL"))/"1 L" = "15.1 meq CaCO"_3"/L"#

Total Alkalinity

Step 1. Calculate the equivalents of #"CaCO"_3#.

#11.80 cancel("mL HCl") × (0.020 cancel("mmol HCl"))/(1 cancel("mL HCl")) × (1cancel("meq HCl"))/(1 cancel("mmol HCl")) × ("1 meq CaCO"_3)/(1 cancel("meq HCl")) = "0.236 meq CaCO"_3#

Step 2. Calculate the total alkalinity

#("0.236 meq CaCO"_3)/(10.0 cancel("mL")) × (1000 cancel("mL"))/"1 L" = "23.6 meq CaCO"_3"/L"#