Question

Given 2 litres of 9.8% w/w ${{H}_{2}}S{{O}_{4}}$ (d = 1.5 gm/ml) solution is mixed with 3 litres of 1 M KOH solution. The concentration of ${{H}^{+}}$ if the solution is acidic or concentration $O{{H}^{-}}$ if the solution is basic in the final solution is?

(A) 0

(B) $\dfrac{3}{10}$

(C) $\dfrac{3}{5}$

(D) $\dfrac{2}{5}$

Answer 1

The basic knowledge of balancing the reaction, its stoichiometry, molarity of the component and idea about neutralisation reactions is necessary to solve the given illustration. Also, only one option can be correct when we solve for physical chemistry illustrations.

Firstly, we need to decide if the solution is acidic or basic before actually solving it which can be made easy by analysing the stoichiometry of the reaction taking place.

Complete answer:

Let us see the basics required to solve this question;

-Balancing the reaction is the primary step and also vital to avoid further chaos.

-We balance the reaction by putting the stoichiometric coefficients before reactants and products.

-Molarity is the concentration of a mole of a component in a litre of solution.

-Neutralisation reaction is the reaction in which acid and base reacts to give salt and water.

Now, keeping these common points in mind let us solve the given problem;

Given data,

Volume of ${{H}_{2}}S{{O}_{4}}$= 2 L = 2000 ml

Density of ${{H}_{2}}S{{O}_{4}}$ = 1.5 g/ml

Thus, mass of ${{H}_{2}}S{{O}_{4}}$ = $2000ml \times 1.5g/ml=3000g$

Weight percent of ${{H}_{2}}S{{O}_{4}}$ in solution = 9.8%

Thus, mass of ${{H}_{2}}S{{O}_{4}}$ in solution = $3000g\times \dfrac{9.8}{100}=294g$

Now, we know molecular weight of ${{H}_{2}}S{{O}_{4}}$ = 98 g/mol

Thus, number of moles of ${{H}_{2}}S{{O}_{4}}$ = $\dfrac{294g}{98g/mol}=3mol$

Molarity of KOH = 1 M

Volume of KOH = 3 L

Thus, number of moles of KOH = $3L\times 1mol/L=3mol$

The neutralisation reaction between ${{H}_{2}}S{{O}_{4}}$ and KOH can be given as,

${{H}_{2}}S{{O}_{4}}+2KOH\to {{K}_{2}}S{{O}_{4}}+2{{H}_{2}}O$

So, by stoichiometry,

2 moles of KOH will neutralise 1 mole of ${{H}_{2}}S{{O}_{4}}$.

Thus, 3 moles of KOH will neutralise 1.5 moles of ${{H}_{2}}S{{O}_{4}}$.

And we have 3 moles of ${{H}_{2}}S{{O}_{4}}$. So, other 1.5 moles will remain un-neutralised.

Thus, number of moles of ${{H}^{+}}$ ions remaining = $2\times$ number of moles of ${{H}_{2}}S{{O}_{4}}$ remaining = 3 moles.

As, we know the total volume of solution = 5 L

Thus, concentration of ${{H}^{+}}$ ions = $\dfrac{3mol}{5L}=\dfrac{3}{5}M$

Therefore, option (C) is correct.

Note: Mendeleev’s periodic table is totally based on the atomic mass of elements. The periodic table we use now is a modern periodic table which is based on the atomic number of elements. Only two Noble gases namely Helium and Argon were discovered by the time of Mendeleev’s periodic table.