# Question b7836

Apr 30, 2015

The phenolphthalein alkalinity is $\text{15.1 meq CaCO"_3"/L}$; the total alkalinity is $\text{23.6 meq CaCO"_3"/L}$.

Alkalinity in water is caused by the presence of ${\text{HCO}}_{3}^{-}$, ${\text{CO}}_{3}^{2 -}$ and ${\text{OH}}^{-}$ 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 ${\text{CO}}_{3}^{2 -}$, ${\text{HCO}}_{3}^{-}$ and ${\text{OH}}^{-}$.

Alkalinity is usually measured in units of $\text{mg CaCO"_3"/L}$ or $\text{meq CaCO"_3"/L}$.

The important concept when working with equivalents is that

$\text{1 eq of anything" = "1 eq of anything else}$.

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 ${\text{CaCO}}_{3}$.

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 ${\text{CaCO}}_{3}$.

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"#