Solveeit Logo
Login

Question

Question: A 100ml of \( 15% \) KOH \( \left( \text{w}/\text{v} \right) \) is used to neutralize 100ml of \( 15...

A 100ml of 1515% KOH (w/v)\left( \text{w}/\text{v} \right) is used to neutralize 100ml of 1515% HCl (w/v)\left( \text{w}/\text{v} \right) . The resultant solution becomes
(A) Alkaline
(B) Acidic
(C) Strongly alkaline
(D) Neutral

Explanation

Solution

Find out the no. of moles for KOH and HCl. It is a neutralization reaction of a strong acid and strong base which includes one mole of base to neutralize one mole of acid. So, on finding the no. of respective moles for acid and base look at the reactant with higher no. of moles this will determine the resultant nature of the solution.

Complete Step by Step Solution
1515% KOH (w/v)\left( \text{w}/\text{v} \right) means that 15g of KOH is present in 100ml of the solution and similarly 1515% HCl (w/v)\left( \text{w}/\text{v} \right) means that 15g of HCl is present in 100ml of the solution, (w/v)\left( \text{w}/\text{v} \right) stands for weight by volume.
No. of moles of KOH =15molar mass of KOH=\dfrac{15}{\text{molar mass of KOH}}
Molar mass of KOH
=39(for potassium)+16(for oxygen)+1(for hydrogen) ⇒39+16+1 ⇒56g \begin{aligned} & =39\left( \text{for potassium} \right)+16\left( \text{for oxygen} \right)+1\left( \text{for hydrogen} \right) \\\ & \Rightarrow 39+16+1 \\\ & \Rightarrow 56\text{g} \\\ \end{aligned}
So, no. of moles of KOH =1556⇒0⋅27=\dfrac{15}{56}\Rightarrow 0\cdot 27
Similarly no. of moles for HCl =1536⋅5⇒0⋅4=\dfrac{15}{36\cdot 5}\Rightarrow 0\cdot 4
The neutralization reaction is: KOH+HCl→KCl+H2O\text{KOH}+\text{HCl}\to \text{KCl}+{{\text{H}}_{2}}\text{O}
Here we can see that one mole of KOH is used to neutralize one mole of HCl. Given no. of moles of HCl are 0⋅40\cdot 4 and for KOH we have no. of moles =0⋅27=0\cdot 27 . If we approximate this to 0⋅30\cdot 3 then we require 0⋅30\cdot 3 moles of KOH to neutralize 0⋅30\cdot 3 moles of HCl. Remaining moles after neutralization are 0⋅4−0⋅3=0⋅10\cdot \text{4}-0\cdot \text{3}=0\cdot 1 mole of HCl. Therefore, the resulting solution will be acidic.
So, the correct choice is (B).

Additional Information
Neutralization reaction is a chemical reaction in which acid and base reacts and forms salt and water. There are strong acids and bases and also there are weak acids and weak bases. Strong acid is the one which dissociates completely in the aqueous solution whereas a weak acid dissociates partially in its aqueous solution. Hydrochloric acid is a strong acid and acetic acid is a weak acid. The neutralization reaction is an exothermic reaction i.e. heat is evolved during this process. Neutralization reactions can take place for any of the combinations like for strong acid and strong base or strong acid and weak base, similarly for weak acid and strong base and other such combinations. If a strong acid and strong base neutralizes fully then the solution becomes neutral, its pH becomes around 7. The enthalpy for such neutralization remains constant i.e. equal to 57⋅1kJ\text{57}\cdot \text{1kJ} because both base and acid undergo complete ionization.

Note
Dissociation constant tells us about the strength of an acid or base as it determines the dissociation of acids and bases into their respective ions in the aqueous solution. The more acid or base dissociates the stronger they are. In the titration process of an acid with a base, the point of neutralization is also called the point of equivalence. Temperature can also affect the resulting pH of the solution.