Question

Question #b7836

Answer 1

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"`