Q4.118CP - Chemistry: Molecular Nature Of Matter And Change

Question

On a lab exam, you have to find the concentrations of the monoprotic (one proton per molecule) acids HA and HB. You are given 43.5mL of HA solution in one flask. A second flask contains 37.2mL of HA, and you add enough HB solution to it to reach a final volume of 50.0mL. You titrate the first HA solution with 87.3mL of 0.0906M NaOH and the mixture of HA and HB in the second flask with 96.4mL of the NaOH solution. Calculate the molarity of the HA and HB solutions.

Step-by-Step Solution

Answer

You need to calculate the molarity of HA and HB solutions.

Step 1: Balanced equations

Titration of HA with NaOH 

 

 $\mathrm{HA}\left(\mathrm{aq}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\to \mathrm{NaA}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

 

Titration of HB with NaOH 

 $\mathrm{HB}\left(\mathrm{aq}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\to \mathrm{NaBr}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 2: Calculation of moles of NaOH

According to the question,

 

Volume of NaOH = 87.3mL = 0.0873L

 

Molarity of NaOH = 0.0906M

 

Again you know,

 

 $\mathrm{moles}=\mathrm{molarity}\times \mathrm{volume}$

 

Now, moles of NaOH

 

 $\begin{array}{l}=(0.0906\times 0.0873)\left(\mathrm{mol}\right)\\ =0.0079\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of NaOH are 0.0079mol.

Step 3: Calculation of moles of HA

According to the balanced equation

 

$\mathrm{HA}\left(\mathrm{aq}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\to \mathrm{NaA}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

 

1 mole of HA reacts with 1 mole NaOH

Now, moles of HA

$\begin{array}{l}=0.0079\times \frac{1}{1}\left(\mathrm{mol}\right)\\ =0.0079\left(\mathrm{mol}\right)\end{array}$

Hence, moles of HA are 0.0079mol.

Step 4: Calculation of molarity of HA

According to the question,

 

Volume of HA = 43.5mL = 0.0435L

 

Again you know,

 

 _{$\mathrm{molarity}=\frac{\mathrm{moles}}{\mathrm{volume}}$}

 

Now, molarity of HA

 

 $\begin{array}{l}=\frac{0.0079}{0.0435}\left(\mathrm{M}\right)\\ =0.1816\left(\mathrm{M}\right)\end{array}$

 

Hence, molarity of HA is 0.1816M.

Step 5: Calculation moles of HA in acid mixture

According to the question,

 

Volume of HA in acid mixture = 37.2mL = 0.0372L

 

Molarity of HA = 0.1816M

 

Again you know,

 

 $\mathrm{moles}=\mathrm{molarity}\times \mathrm{volume}$

 

Now, moles of HA in acid mixture

 

 $\begin{array}{l}=(0.1816\times 0.0372)\left(\mathrm{mol}\right)\\ =0.00676\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of HA in acid mixture are 0.00676mol.

 

Step 6: Calculation of volume of NaOH for titration of HA in acid mixture

As moles of HA in acid mixture are 0.00676mol, therefore moles of NaOH for titration of HA in acid mixture is also 0.00676mol.

 

Again you know,

 

 $\mathrm{volume}=\frac{\mathrm{moles}}{\mathrm{molarity}}$

 

Now, volume of NaOH for titration of HA in acid mixture

 $\begin{array}{l}=\frac{0.00676}{0.0906}\left(\mathrm{L}\right)\\ =0.07461\left(\mathrm{L}\right)\\ =74.61\left(\mathrm{mL}\right)\end{array}$

 

 

Hence, volume of NaOH for titration of HA in acid mixture is 74.61mL.

Step 7: Calculation of volume of NaOH for titration of HB in acid mixture

Now, volume of NaOH for titration of HB in acid mixture

 $\begin{array}{l}=(96.4-74.61)\left(\mathrm{mL}\right)\\ =21.79\left(\mathrm{mL}\right)\end{array}$

 

 

Hence, volume of NaOH for titration of HB in acid mixture is 74.61mL.

Step 8: Calculation of moles of NaOH for titration of HB in acid mixture

Volume of NaOH for titration of HB in acid mixture = 21.79mL = 0.02179L

 

Molarity of NaOH = 0.0906M

 

Again you know,

 

 $\mathrm{moles}=\mathrm{molarity}\times \mathrm{volume}$

 

Now, moles of NaOH for titration of HB in acid mixture

 $\begin{array}{l}=(0.0906\times 0.02179)\left(\mathrm{mol}\right)\\ =0.001974\left(\mathrm{mol}\right)\end{array}$

 

 

Hence, moles of NaOH for titration of HB in acid mixture are 0.00676mol.

Step 9: Calculation of moles of HB in acid mixture

According to the balanced equation

 

$\mathrm{HBr}\left(\mathrm{aq}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\to \mathrm{NaBr}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

 

1 mole of HB reacts with 1 mole NaOH

 

Now, moles of HB in acid mixture

 

 $\begin{array}{l}=0.001974\times \frac{1}{1}\left(\mathrm{mol}\right)\\ =0.001974\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of HB in acid mixture are 0.001974mol.

Step 10: Calculation of volume of HB in acid mixture

Now, volume of HB in acid mixture

 $\begin{array}{l}=\mathrm{volume}\left(\mathrm{mixture}\right)-\mathrm{volume}\left(\mathrm{HA}\right)\\ =(50-37.2)\left(\mathrm{mL}\right)\\ =12.8\left(\mathrm{mL}\right)\end{array}$

 

 

Hence, volume of HB in acid mixture is 12.8mL.

Step 11: Calculation of molarity of HB in acid mixture

Volume of HB in acid mixture = 12.8mL = 0.0128L

 

Again you know,

 

 _{$\mathrm{molarity}=\frac{\mathrm{moles}}{\mathrm{volume}}$}

 

Now, molarity of HB in acid mixture

 

 $\begin{array}{l}=\frac{0.001974}{0.0128}\left(\mathrm{M}\right)\\ =0.154\left(\mathrm{M}\right)\end{array}$

 

Hence, molarity of HB in acid mixture is 0.154M.

Step 12: Conclusion

Molarity of HA is 0.1816M and molarity of HB is 0.154M.