# How to calculate the concentrations of the species in the carbonate equilibrium from a titration with hydrochloric acid? [closed]

Lake water contains dissolved sodium carbonate and sodium hydrogen carbonate. The following equilibrium exists: \begin{align} \ce{HCO3- &<=> H+ + CO3^{2-}}& \frac{\ce{[CO3^{2-}]}}{\ce{[HCO3-]}} &= 0.958 \end{align} When $$\pu{10 cm^3}$$ of lake water were titrated with $$\pu{0.2 mol/dm^3}$$ $$\ce{HCl}$$, $$\pu{22 cm^3}$$ of acid were required to neutralise all the carbonate and hydrogen carbonate ions according to the following equations: \begin{align} \ce{H+ + HCO3- &-> H2O + CO2}\\ \ce{2H+ + CO3^{2-} &-> H2O + CO2}\\ \end{align} Calculate the total amount of substance of acid used, and thus, by using the ratio quoted, calculate $$\ce{[CO3^{2-}]}$$ and $$\ce{[HCO3-]}$$ in the lake.

I was able to find the amount of substance of acid $$\ce{HCl}$$, which is $$\pu{4.4E-3 mol}.$$ How do I go on?

## closed as off-topic by Klaus-Dieter Warzecha, Wildcat, M.A.R., getafix, Todd MinehardtOct 31 '16 at 13:01

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You have correctly identified the amount of substance $$\ce{HCl}$$ which were necessary to neutralize all $$\ce{HCO3-}$$ and $$\ce{CO3^2-}$$ in the sample. With the given sample volume of $$\pu{10 cm^3}$$, we can calculate the concentration of both ions in the sample: $$\frac{\pu{4.4E-3 mol}}{\pu{10 cm^3}} = \pu{4.4E-4 mol//cm^3} = [\ce{HCO3-}]+[\ce{CO3^{2-}}]$$
Using the given ratio of $$0.958$$, we can express $$[\ce{CO3^{2-}}]$$ in terms of $$[\ce{HCO3^{-}}]$$: $$\begin{gather} [\ce{CO3^2-}] = 0.958\times[\ce{HCO3-}]\\ \pu{4.4E-4 mol//cm^3} = [\ce{HCO3-}] + [\ce{CO3^2-}] = [\ce{HCO3-}] + 0.958\times[\ce{HCO3-}] = 1.958[\ce{HCO3-}]\\ [\ce{HCO3-}] = \pu{2.25E-4 mol//cm3} \end{gather}$$
We then use the calculated hydrogen carbonate concentration and the ratio to find $$[\ce{CO3^2-}]$$: $$[\ce{CO3^2-}] = 0.958\times[\ce{HCO3-}] = \pu{2.15E-4 mol//cm3}$$