Q4.115CP - Chemistry: Molecular Nature Of Matter And Change

Question

You are given solutions of HCl and NaOH and must determine their concentrations. You use 27.5mL of NaOH to titrate 100mL of HCl and 18.4mL of NaOH to titrate 50.0mL of 0.0782M H₂SO₄. Find the unknown concentrations.

Step-by-Step Solution

Answer

Unknown concentrations mentioned in the question should be calculated.

Step 1: Balanced equation of NaOH – H 2 SO 4 titration

Titration of sulfuric acid by sodium hydroxide produces sodium sulfate and water. The balanced equation is like

 

 $2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 2: Calculation of moles of H 2 SO 4

According to the question,

 

Volume of H₂SO₄ = 50mL = 0.05L

 

Molarity of H₂SO₄ = 0.0782M

 

Again you know,

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

 

 

Now, moles of H₂SO₄

 $\begin{array}{l}=(0.0782\times 0.05)\left(\mathrm{mol}\right)\\ =0.00391\left(\mathrm{mol}\right)\end{array}$

 

 

Hence, moles of H₂SO₄ are 0.00391mol.

Step 3: Calculation of moles of NaOH

According to the balanced equation

 

$2\mathrm{NaOH}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

 

2 moles of NaOH react  with 1 mole of H₂SO₄

 

Now, moles of NaOH

$\begin{array}{l}\mathbf{=}\mathbf{0}\mathbf{.00391}\mathbf{\times }\frac{\mathbf{2}}{\mathbf{1}}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\\ \mathbf{=}\mathbf{0}\mathbf{.00782}\mathbf{\left(}\mathbf{mol}\mathbf{\right)}\end{array}$

Hence, moles of H₂SO₄ are 0.00782mol.

Step 4: Calculation of molarity of NaOH

According to the question,

 

Volume of NaOH = 18.4mL = 0.0184L

 

Again you know,

 

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

 

Now, molarity of NaOH

 

 $\begin{array}{l}=\frac{0.00782}{0.0184}\left(\mathrm{M}\right)\\ =0.425\left(\mathrm{M}\right)\end{array}$

 

Hence, molarity of NaOH is 0.425M.

Step 5: Balanced equation for NaOH - HCl titration

The balanced equation is 

 

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

Step 6: Calculation of moles of NaOH is required to titrate HCl

According to the question,

 

Volume of NaOH required = 27.5mL = 0.0275L

 

Molarity of NaOH calculated = 0.425M

 

Again you know,

 

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

 

Now, moles of NaOH

 

 $\begin{array}{l}=(0.425\times 0.0275)\left(\mathrm{mol}\right)\\ =0.01169\left(\mathrm{mol}\right)\end{array}$

 

Hence, moles of NaOH required to titrate HCl are 0.01169mol.

Step 7: Calculation of moles of HCl

According to the balanced equation

 

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

 

2 moles of NaOH reacts with 1 mole of HCl

 

Now, moles of HCl

 ^{$\begin{array}{l}=0.01169\times \frac{1}{1}\left(\mathrm{mol}\right)\\ =0.01169\left(\mathrm{mol}\right)\end{array}$}

 

 

Hence, moles of HCl are 0.01169mol.

Step 8: Calculation of molarity of HCl

According to the question,

 

Volume of HCl = 100mL = 0.1L

 

Again you know,

 

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

 

Now, molarity of HCl

 

 $\begin{array}{l}=\frac{0.01169}{0.1}\left(\mathrm{M}\right)\\ =0.1169\left(\mathrm{M}\right)\end{array}$

 

Hence, molarity of HCl is 0.1169M.

Step 9: Conclusion

Hence, molarity of NaOH is 0.425M and molarity of HCl is 0.1169M.