Question

A -$100mL$solution $0.1N$ $HCl$ was titrated with $0.2N$ $NaOH$ a solution. The titration was discontinued after adding $30mL$ of $NaOH$ solution. The titration was completed by adding $0.25NKOH$ a solution. The volume of $KOH$ required for the complete titration is:
$70mL$
$32mL$
$35mL$
$16mL$

Answer 1

Hint: Titration is the process of chemical analysis in which the concentration of a constituent of a sample can be evaluated by the known concentration of the other substituent. The concentration of an acidic solution can be evaluated by titrating it with a standard basic solution of known concentration required to neutralize it.

Step-by-step explanation:
Step 1:
As we know that for a complete neutralization process the number of gram equivalence of acid is always equal to the number of gram equivalence of the base.
The number of gram equivalence can be calculated by two methods:
$number\,of\,gm\,equivalence = \dfrac{{weight\,of\,the\,solute\,given}}{{equivalent\,weight\,of\,the\,solute}}$
Again we know that the number of gram equivalents of the solute dissolved in a liter of solution at a particular temperature is called the normality of the solution.
$Normality = \dfrac{{Number\,of\,gram\,equivalence\,of\,the\,solute}}{{Volume\,of\,solution\,in\,L}}$

$Number\,of\,gram\,equivalence = Normality \times Volume$
Step 2:
$HCl + NaOH\xrightarrow{{}}NaCl + {H_2}O$
So, for complete neutralization, the one-gram equivalence of HCl will react with the one-gram equivalence of NaOH.
Step 3:
As we know that $Number\,of\,gram\,equivalence = Normality \times Volume$
So, the number of gram equivalence for $100mL$ solution of $0.1N$ $HCl$is: [ N is the concentration in normality]
$0.1N \times 0.1L = 0.01gequivalence$ [As $100mL = 0.1L$] [L stands for litre]
In the same way, the number of gram equivalence for 30mLof0.2N$$$$NaOHis:
$0.2N \times 0.03L = 0.006gequivalence$ [ As $30mL = 0.03L$]
So, 0.006 gm equivalence of HCl was neutralized with NaOH.

Step 4:
As the number of gram equivalents of HCl is 0.01 whereas for NaOH is 0.006 so complete neutralization process has not taken place.
The number of gram equivalents of HCl remaining back in the solution is:
$0.01 - 0.006 = 0.004$
Step 5:
So, 0.004 gm equivalence of HCl has to be neutralized using $0.25NKOH$.
So, the volume of KOH required $= \dfrac{{Number\,of\,gm\,equivalence}}{{Normality}}$
Volume of KOH $= \dfrac{{0.004}}{{0.25}}$
As $Normality = \dfrac{{no.of\,gm\,equivalence}}{{volumeinL}}$
The volume of KOH (L) $= 0.016$L
The volume of KOH required $16mL$ [$1L = 1000mL$]

Hence, the volume of KOH required to complete the process of neutralization is $16mL$.

Note: For complete neutralization in the process of titration the number of gram equivalence of the acid is always equal to the number of gram equivalence of a base.