Chapter 15, Problem 15.106QA

Interpretation Introduction

To find:

The concentration of carbonate $\left(C{O}_3^{2-}\right)$ and bicarbonate $\left(HC{O}_3^{-}\right)$ ions in the sample of water from a hot spring.

Answer to Problem 15.106QA

Solution:

The concentration of carbonate ions in the water sample is $9.09 \times {10}^{-4}M,$ and that of bicarbonate ions is $2.79\times {10}^{-3}M$.

Explanation of Solution

1) Concept:

We are asked to find concentrations of carbonate $\left(C{O}_3^{2-}\right)$ and bicarbonate $\left(HC{O}_3^{-}\right)$ ions in the sample of water from the results of a titration. These determinations are based on the volumes of titrant needed to reach two equivalence points.

In the first stage, after the addition of the titrant, the carbonate ions present in the sample are converted into bicarbonate ions at the first equivalence point.

$C{O}_3^{2-} \left(aq\right)+H_3{O}^{+}\left(aq\right)\to HC{O}_3^{-} \left(aq\right)+H_{2}O \left(l\right)$

The reaction continues until all the carbonate ions are has been converted into bicarbonate ions. This is the first equivalence point in the titration. In the second stage of titration, the bicarbonate ions formed in the first stage plus any bicarbonate ions present in the original sample react with the additional titrant, forming carbonic acid,$H_{2}C{O}_3\left(aq\right)$.

$HC{O}_3^{-} \left(aq\right)+H_3{O}^{+}\left(aq\right)\to H_{2}C{O}_3\left(aq\right)+H_{2}O \left(l\right)$

According to the stoichiometry of the reactions, it takes 1 mole of $HCl$ to titrate 1 mole of carbonate to bicarbonate in the first stage, and it takes 1 mole of $HCl$ to titrate 1 mol of bicarbonate to carbonic acid in the second stage.

2) Formula:

$M=\frac{mol solute}{L solution}$

3) Given:

i) Volume of sample = $100.0 mL$

ii) Volume of $HCl$ required to reach the first equivalence point at $pH=8.3$= $2.56 mL$

iii) Molarity of $HCl$ used = $0.0355 M$

iv) Volume of $HCl$ required to reach the second equivalence point at $pH=4.0$= $10.42-2.56=7.86 mL$

v) The alkalinity of water is due to carbonate $\left(C{O}_3^{2-}\right)$ and bicarbonate $\left(HC{O}_3^{-}\right)$ ions.

4) Calculations:

Calculating concentrations of carbonate $\left(C{O}_3^{2-}\right)$ and bicarbonate $\left(HC{O}_3^{-}\right)$ ions in the original sample from the volumes and molarity of $HCl$ titrant consumed in stage 1 and 2:

$\left[{CO}_3^{2-}\right]=\frac{mol {CO}_3^{2-}}{L solution}$

$Moles of carbonate=\left(2.56 mL HCl\times \frac{1 L}{1000 mL}\times \frac{0.0355 mol}{1 L HCl} \times \frac{1 mol {CO}_3^{2-}}{1 mol HCl}\right)=9.09\times {10}^{-5} mol$

 Therefore, the concentration of carbonate is

$\left[{CO}_3^{2-}\right]=\frac{9.09\times {10}^{-5} mol}{100.0 mL solution \times \frac{1L}{1000 mL}}$

$=9.09 \times {10}^{-4} M$

These moles of carbonate are converted to an equal number of moles of bicarbonate at the first equivalence point. The solution now contains these moles plus the moles of bicarbonate already present in the spring water sample. These correspond to the titration from the first to the second equivalence point, i.e., to the addition of another 10.42 mL of the acid. The total bicarbonate moles are then calculated as

$Total moles of bicarbonate=\left(10.42 mL HCl\times \frac{1 L}{1000 mL}\times \frac{0.0355 mol}{1 L HCl} \times \frac{1 mol {CO}_3^{2-}}{1 mol HCl}\right)=3.70\times {10}^{-4} mol$

Of these, $9.09\times {10}^{-5}$ moles are from the first part of the titration, so their difference will give us the moles of bicarbonate originally present in the sample.

$moles of bicarbonate in sample=total moles of bicarbonate-moles of bicarbonate formed in first part of tiration=3.70\times {10}^{-4}mol-9.09\times {10}^{-5}mol=2.79\times {10}^{-4} mol$

The concentration of bicarbonate in the sample is then calculated as

$\left[{HCO}_3^{-}\right]=\frac{mol {HCO}_3^{-}}{L solution}=\frac{2.79\times {10}^{-4} mol}{100 mL\times \frac{1 L}{1000 mL}}= 2.79 \times {10}^{-3} M$

Conclusion:

The volume required for the second equivalent point is three times that for the first equivalence point. This is consistent with the observed concentrations of carbonate and bicarbonate ions as the concentration of bicarbonate ions is three times that of the carbonate ions.