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.

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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