Question

What volume of $ 0.010M $ $ {{H}_{2}}S{{O}_{4}} $ must be added to $ ~50mL $ of a $ 0.10M $ $ NaOH $ solution to make a solution in which the molarity of the $ {{H}_{2}}S{{O}_{4}} $ is $ 0.050M $ ?
(A) $ 400mL $
(B) $ 50mL $
(C) $ 100mL $
(D) $ 150mL $

Answer 1

Hint :We know that molarity is defined as the measure of the concentration. It is expressed as the number of moles of solute per litre of the solution. Molarity $ \left( M \right) $ determines the number of moles of solute per litre of solution which is denoted as $ moles/Liter $ and is amongst the most common units that are used for measuring the concentration of a solution.

Complete Step By Step Answer:
The answer to the question lies in the concept of dilution. A dilution of a solution occurs when a solution of a certain concentration is added more solvent in which the substance under consideration is likely to dissolve. This results in the decrease of concentration with respect to the original solution. By adding more solvent to the solution the total volume of the solution increases as the concentration of the solution decreases.
If we add $ \,V\text{ }ml $ of $ 0.1\text{ }M $ $ {{H}_{2}}S{{O}_{4}} $ ​ then you are adding $ 0.01 $ moles of $ {{H}_{2}}S{{O}_{4}} $ .
In $ 50mL $ of $ 0.1\text{ }M\text{ }NaOH $ , there are $ 0.005 $ moles of $ NaOH. $
The chemical equation for that is: $ 2NaOH+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O. $
 $ 0.0025 $ moles of $ {{H}_{2}}S{{O}_{4}} $ ​ will neutralize $ 0.005 $ moles of $ NaOH $ . So from $ 0.01V $ moles of $ {{H}_{2}}S{{O}_{4}} $ ​ added in $ VmL $ there will be $ 0.01V-\text{ }0.0025 $ moles of $ {{H}_{2}}S{{O}_{4}} $ ​ left unused in the total volume of $ V\text{ }+\text{ }50mL $ of solutions.
Thus, the concentration will be $ \dfrac{0.01V-0.0025}{V+50}\times 1000=0.05M $
On further solving we get;
 $ \dfrac{100V-2.5}{V+50}=0.05 $
For solving $ V $ , we get the values as;
 $ V=0.05002L $ or $ 50.02mL\text{ }=\text{ }50mL $
Therefore correct answer is option B, i.e. $ 50mL $

Note :
Remember that the boiling point of haloalkanes decreases with branching. The haloalkanes are only very slightly soluble in water, but dissolves readily in organic solvents. The concentration and the volume of the concentrated or dilute solution can be determined using the equation of dilution: $ {{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}. $