Question

A student added 50.0 mL of an NaOH solution to 100.0 mL of 0.400 M HCl. The solution was then treated with an excess of aqueous chromium (III) nitrate, resulting in formation of 2.06 g of precipitate. Determine the concentration of the NaOH solution

Short answer

Thus the molarity of NaOH solution is 2 M.

Step-by-step solution

Definition of molarity of a solution.

The concentration of the solution also called as molarity is basically the concentration can be calculated by fractionating the number of moles with the volume of solution in litre.

It can be formulated as

$\text{Molarity}\left(M{\text{ormolL}}^{-1}\right)=\frac{\text{Numberofmoles}\left(n\right)}{\text{Volumeofsolution}\left(\text{inL}\right)}\cdots \cdots \left(1\right)$

The neutralization reaction of sodium hydroxide and hydrochloric acid. 

The neutralization reaction the given reactants result in production of sodium chloride along with elimination of water and the reaction can be written as

$\text{NaOH}+\text{HCl}\to {\text{H}}_{\text{2}}\text{O}+\text{NaCl}$

Data given in the question and conversion of volume from mL to L. 

Data given in the question,

• Initial molarity = 0.400 M
• Final molarity =?
• Initial volume = 50 mL
• Final volume = 100 mL

Since in formula of molarity the volume of solution is present in L thus 50 mL can be converted to L as

$\begin{array}{l}\text{1L=1000mL}\\ \text{Conversionfactor=}\frac{\text{1L}}{\text{1000mL}}\\ \text{Volume}\left(\text{inL}\right)\text{=Volume}\left(\text{inmL}\right)\times \text{Conversionfactor}\\ \text{Volume}\left(\text{inL}\right)\text{=50}\bcancel{\text{mL}}\times \frac{\text{1L}}{\text{1000}\bcancel{\text{mL}}}\\ =\text{0.05L}\end{array}$

The final volume of solution is present in L thus 100 mL can be converted to L as

$\begin{array}{l}\text{1L=1000mL}\\ \text{Conversionfactor=}\frac{\text{1L}}{\text{1000mL}}\\ \text{Volume}\left(\text{inL}\right)\text{=Volume}\left(\text{inmL}\right)\times \text{Conversionfactor}\\ \text{Volume}\left(\text{inL}\right)\text{=100}\bcancel{\text{mL}}\times \frac{\text{1L}}{\text{1000}\bcancel{\text{mL}}}\\ =\text{0.1L}\end{array}$

Calculating the number of moles of NaOH and HCl in solution.

The number of moles of the solution can be calculated by rearranging equation (1) as

$\begin{array}{c}\text{Numberofmoles}\left(n\right)=\text{Molarity}\left({\text{molL}}^{-\text{1}}\right)\times \text{Volumeofsolution}\left(\text{inL}\right)\\ \text{=}\frac{\text{0.4molHCl}}{\bcancel{\text{L}}}\times 0.100\bcancel{\text{L}}\text{HCl}\\ =0.04\text{molHCl}\end{array}$

The number of moles remain same throughout the reaction, the number of moles of NaOH will also be 0.04 mol NaOH.

Precipitation reaction of sodium hydroxide and chromium nitrate.

The number of moles of the solution can be calculated by rearranging equation (1) as $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}+\text{3NaOH}\to \text{Cr}{\left(\text{OH}\right)}_{\text{3}}+{\text{3NaNO}}_{\text{3}}$

Molar mass of chromium (III) hydroxide.

The molar mass can be calculated as

$\begin{array}{c}\text{MolarmassofCr}{\left(\text{OH}\right)}_{\text{3}}=\text{AtomicmassofCr}+3\left(\text{AtomicmassofO}\right)+3\left(\text{AtomicmassofH}\right)\\ =52{\text{gmol}}^{-1}+3\left(16{\text{gmol}}^{-1}\right)+3\left(1{\text{gmol}}^{-1}\right)\\ =52{\text{gmol}}^{-1}+48{\text{gmol}}^{-1}+3{\text{gmol}}^{-1}\\ =103{\text{gmol}}^{-1}\end{array}$

Calculating the number of moles of NaOH in solution. 

The number of moles of the solution can be calculated as

$\begin{array}{c}\text{Numberofmoles}\left(n\right)=\frac{\text{Givenmass}}{\text{Molarmass}}\\ \text{=}\left({\text{2.06gCrOH}}_{\text{3}}\right)\left(\frac{\text{1molCr}{\left(\text{OH}\right)}_{\text{3}}}{103{\text{gmol}}^{-1}}\right)\left(\frac{\text{3molOH}}{\text{1moleCr}}\right)\\ =0.06\text{molHCl}\end{array}$

Total number of moles of NaOH in solution. 

The total number of moles of the solution can be calculated as

$\begin{array}{c}\text{TotalnumberofNaOH}=0.04\text{molNaOH}+0.06\text{molNaOH}\\ \text{=0.10molNaOH}\end{array}$

Calculating the molarity of NaOH solution. 

The molarity of the solution can be calculated by rearranging equation (1) as

$\begin{array}{l}\text{Molarity}\left({\text{molL}}^{-\text{1}}\right)\text{=}\frac{\text{Numberofmoles}\left(n\right)}{\text{Volumeofsolution}\left(\text{inL}\right)}\\ \text{=}\frac{\text{0.10molNaOH}}{\text{0.050L}}\\ =2.0{\text{molL}}^{-\text{1}}\text{NaOH}\\ =2.0\text{MNaOH}\end{array}$

Thus the molarity of NaOH solution is 2 M.