Question

What is the molarity of sodium hydroxide solution?
A 0.3178g sample of potassium hydrogen sulphate is dissolved in 100 mL of water. 33.75 mL of sodium hydroxide solution is required to reach the equivalence point.

Answer 1

Hint: As we know that molarity is the number of moles of a solute in 1 litre of solution so here we need to calculate molarity of sodium hydroxide in 33.75 mL of solution. Also the equivalence point is the point in a chemical reaction where exactly enough acid and base is present to neutralize each other.
Formula used:
We will use the following formulas in this solution:-
$M=\dfrac{n}{V}$
where,
M = molarity
n = number of moles of solute.
V = volume in liters
Also at equivalence point: ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$

Complete answer:
Let us first understand about molarity and then solve this question as follows:-
-We know that molarity is a concentration term which is the ratio of number of moles of a solute in 1 litre of solution. Mathematically it is shown as follows:-
$M=\dfrac{n}{V}$
where,
M = molarity
n = number of moles of solute.
V = volume in liters
-Calculation of number of moles of potassium hydrogen sulphate (${{C}_{8}}{{H}_{5}}K{{O}_{4}}$):-
As we know that potassium hydrogen sulphate (${{C}_{8}}{{H}_{5}}K{{O}_{4}}$) reacted with sodium hydroxide (NaOH) in such a way that they reach exactly at equivalence point, so first we will calculate the molarity of potassium hydrogen sulphate.
Given that 0.3178g of sample of potassium hydrogen sulphate (${{C}_{8}}{{H}_{5}}K{{O}_{4}}$) is dissolved in 100 mL of water (i.e., 0.1L). So number of moles of ${{C}_{8}}{{H}_{5}}K{{O}_{4}}$= $\dfrac{\text{given mass of }{{C}_{8}}{{H}_{5}}K{{O}_{4}}}{\text{molecular mass of }{{C}_{8}}{{H}_{5}}K{{O}_{4}}}$
Given mass of ${{C}_{8}}{{H}_{5}}K{{O}_{4}}$ = 0.3178g
Molecular mass of ${{C}_{8}}{{H}_{5}}K{{O}_{4}}$= 8(12) +5(1) + 39 + 4(16) = 204g/mol
So number of moles of ${{C}_{8}}{{H}_{5}}K{{O}_{4}}$= $\dfrac{0.3178g}{204g/mol}=0.00156mol$
Molarity of${{C}_{8}}{{H}_{5}}K{{O}_{4}}$: $\dfrac{0.00156mol}{0.1L}=0.0156M$
-Calculation of molarity of sodium hydroxide solution using equivalence point:-
As we know that equivalence point (also known as stoichiometric point) is the point in a reaction where exactly enough acid and base is present to neutralize each other. It is where the moles of titrant are equal to the moles of solution of unknown concentration in a titration.
So if ${{n}_{1}}={{n}_{2}}$then${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$ as n = MV.
So we will apply the same for the reaction of ${{C}_{8}}{{H}_{5}}K{{O}_{4}}$with NaOH as follows:-
${{M}_{1}}$= Molarity of ${{C}_{8}}{{H}_{5}}K{{O}_{4}}$= 0.0156M
${{V}_{1}}$= Volume of ${{C}_{8}}{{H}_{5}}K{{O}_{4}}$= 100mL
${{V}_{2}}$= Volume of NaOH= 33.75mL
On putting all the values in equation${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$, we get:-
$\begin{align}
  & \Rightarrow {{M}_{2}}=\dfrac{{{M}_{1}}{{V}_{1}}}{{{V}_{2}}} \\
 & \Rightarrow {{M}_{2}}=\dfrac{0.0156M\times 100mL}{33.75mL} \\
 & \Rightarrow {{M}_{2}}=0.0462M \\
\end{align}$
-Therefore the molarity of sodium hydroxide solution is = 0.0462M

Note:
-Always remember to convert units according to the requirement of the formulas being used and also prefer to do calculations along with the units.
-No need to convert units in case of${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$, as unit of volumes get canceled whether it is in millilitres or litres.