Chapter 15, Problem 15.108QA

Interpretation Introduction

To find:

What volume of $0.0100 M HCl$  is required to titrate $250 mL of 0.0100 M {Na}_{2}C{O}_3$ and $250 mL$ of $0.0100 M HC{O}_3^{-}$?

Answer to Problem 15.108QA

Solution:

a) $500.0 mL 0.100 M HCl$ is required to titrate with $250 mL of 0.0100 M {Na}_{2}C{O}_3$.

b) $250.0 mL 0.100 M HCl is \mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{ }\mathrm{t}\mathrm{o}\mathrm{ }\mathrm{t}\mathrm{i}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{ }\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\mathrm{ }250 mL of 0.0100 M HC{O}_3^{-}.$

Explanation of Solution

1) Concept:

We are asked to find the volumes of $HCl$ that are required to neutralize $NaHC{O}_3$ and $N{a}_{2}C{O}_3$ solutions of known molarity and volume.

The complete neutralization equation for the titration of $HCl$ and ${Na}_{2}C{O}_3$ is:

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

From the stoichiometry of the reaction, it can be seen that one mole of $C{O}_3^{2-}$ requires two moles of $H^{+}$ ions. We have equal molarity values for carbonate and $HCl$, so it can be estimated that, double volume of $HCl$ will be required to neutralize $250 mL$ solution of carbonate ions.

Similarly, the complete neutralization equation for the titration of $HCl$ and $NaHC{O}_3$ is:

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

From the stoichiometry of the reaction, it can be seen that one mole of $HC{O}_3^{-}$ requires one mole of $H^{+}$ ions to neutralize completely. We have equal molarity values for carbonate and $HCl$, so it can be estimated that the volume of $HCl$ that will be required to neutralize $250 mL$ solution of carbonate ions is $250.0 mL$.

2) Formula:

$n=\frac{M}{V}$

3) Given:

i) Molarity of ${Na}_{2}C{O}_3=0.0100 M$

ii) Volume of ${Na}_{2}C{O}_3=250 mL$

iii) Molarity of $HC{O}_3^{-}=0.0100 M$

iv) Volume of $HC{O}_3^{-}=250 mL$

v) Molarity of $HCl=0.0100 M$

4) Calculation:

a) The volume of $0.0100 M HCl$  is required to titrate $250 mL of 0.0100 M {Na}_{2}C{O}_3$

Converting the volume in mL to L

$250 mL {Na}_{2}C{O}_3\times \frac{1 L}{1000 mL}=0.250 L {Na}_{2}C{O}_3$

Calculating the moles of ${Na}_{2}C{O}_3$

$Mol of {Na}_{2}C{O}_3=\frac{0.0100 mol}{1 L}\times 0.250 L=0.00250 mol$

$Mol of {Na}_{2}C{O}_3=mol of C{O}_3^{2-}\left(aq\right)=0.00250 mol$

The reaction for reaction titration of $HCl$ and ${Na}_{2}C{O}_3$ is,

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

The mole ratio of $C{O}_3^{2-}:H^{+}$ is 1:2

Therefore, moles of $H^{+}$ i.e. $HCl$ required are

$0.00250 mol C{O}_3^{2-}\times \frac{2 mol H^{+} }{1 mol C{O}_3^{2-}}=0.0050 mol H^{+}$

i.e. $Moles of HCl=0.0050 mol$

Calculating the volume of $HCl$,

$0.0050 mol HCl\times \frac{1 L}{0.0100 mol}=0.50 L HCl$

Converting volume in L to mL

$0.50 L HCl\times \frac{1000 mL}{1 L}=500.0 mL HCl$

Therefore, $500.0 mL 0.100 M HCl$ is required to titrate with $250 mL of 0.0100 M {Na}_{2}C{O}_3$.

b) The volume of $0.0100 M HCl$  required to titrate $250 mL of 0.0100 M HC{O}_3^{-}$

Converting the volume in mL to L

$250 mL HC{O}_3^{-}\times \frac{1 L}{1000 mL}=0.250 L HC{O}_3^{-}$

Calculating the moles of $HC{O}_3^{-}$

$Mol of HC{O}_3^{-}=\frac{0.0100 mol}{1 L}\times 0.250 L=0.00250 mol$

The reaction for titration of $HCl$ and $HC{O}_3^{-}$ is

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

The mole ratio of $HC{O}_3^{-}:H^{+}$ is 1:1

Therefore, moles of $H^{+}$ i.e. $HCl$ required are

$0.00250 mol HC{O}_3^{-}\times \frac{1 mol H^{+} }{1 mol HC{O}_3^{-}}=0.00250 mol H^{+}$

i.e. $Moles of HCl=0.000250 mol$

Calculating the volume of $HCl$

$0.00250 mol HCl\times \frac{1 L}{0.0100 mol}=0.250 L HCl$

Converting volume in L to mL

$0.250 L HCl\times \frac{1000 mL}{1 L}=250.0 mL HCl$

Therefore, $250.0 mL 0.100 M HCl s required to titrate with 250 mL of 0.0100 M HC{O}_3^{-}.$

Conclusion:

Stoichiometry of the reaction of carbonate and bicarbonate ions with a strong acid such as $HCl$ can be used to find the volume of $HCl$ required to reach the equivalence points.