Question

When 100mL of 0.1M $Ba{\left( {OH} \right)_2}$ is neutralized with a mixture of x mL of 0.1M HCl and y mL of 0.2M ${H_2}S{O_3}$ using methyl orange indicator, What is the value of x and y respectively?
a.) 200, 100
b.) 100, 200
c.) 300, 200
d.) 200, 300

Answer 1

Hint: We will solve the solution by dividing it into two cases where we will get to know that the normality and molality will be the same. Thus, applying the formula ${N_1}{V_1} = {N_2}{V_2}$, we will get the values of x and y respectively. Refer to the solution below.
Complete answer:
 ${N_1}{V_1} = {N_2}{V_2}$, $N = M \times n$.
To find out- the volume of HCl and ${H_2}S{O_3}$.
The volume of HCl is given as – x
The volume of ${H_2}S{O_3}$ is given as – y
The fact about methyl orange is that this indicator indicates only when the reaction is a hundred percent complete.
The reaction of ${H_2}S{O_3}$ will occur as-
$ \Rightarrow {H_2}S{O_3} \to {H^ + } + HS{O_3}^ - $
Thus, it will be ionized. Where, the n factor of the above reaction will be 1 and the normality will be n factor multiplied by molality-
$
   \Rightarrow n = 1 \\
    \\
   \Rightarrow N = M \times n \\
    \\
   \Rightarrow N = M \\
$
Case 1-
$
   \Rightarrow Ba{\left( {OH} \right)_2} \equiv HCl \\
    \\
   \Rightarrow {N_1}{V_1} = {N_2}{V_2} \\
$
Where, it is given that-
$
   \Rightarrow {N_1} = M \times n \\
    \\
   \Rightarrow {N_1} = 0.1 \times 2 \\
    \\
   \Rightarrow {V_1} = 100mL \\
$
And,
$
   \Rightarrow {N_2} = M \times n \\
    \\
   \Rightarrow {N_2} = 0.1 \times 1 \\
    \\
   \Rightarrow {V_2} = xmL \\
$
Substituting these values in the above equation ${N_1}{V_1} = {N_2}{V_2}$, we get-
$
   \Rightarrow {N_1}{V_1} = {N_2}{V_2} \\
    \\
   \Rightarrow 0.1 \times 2 \times 100 = x \times 0.1 \times 1 \\
    \\
   \Rightarrow x = 200mL \\
$
Case 2-
$
   \Rightarrow Ba{\left( {OH} \right)_2} \equiv {H_2}S{O_3} \\
    \\
   \Rightarrow {N_1}{V_1} = {N_2}{V_2} \\
$
Where, it is given that-
$
   \Rightarrow {N_1} = M \times n \\
    \\
   \Rightarrow {N_1} = 0.1 \times 2 \\
    \\
   \Rightarrow {V_1} = 100mL \\
$
And,
$
   \Rightarrow {N_2} = M \times n \\
    \\
   \Rightarrow {N_2} = 0.2 \times 1 \\
    \\
   \Rightarrow {V_2} = ymL \\
$
Substituting these values in the above equation ${N_1}{V_1} = {N_2}{V_2}$, we get-
$
   \Rightarrow {N_1}{V_1} = {N_2}{V_2} \\
    \\
   \Rightarrow 0.1 \times 2 \times 100 = y \times 0.2 \times 1 \\
    \\
   \Rightarrow y = 100mL \\
$
Hence, it is clear that option A is the correct option.

Note: Methyl orange, because of its simple and distinct colour variation at various pH levels, is a pH indicator commonly used in titration. In acidic medium and yellow in basic medium Methyl Orange is a red pigment. As the $p{K_a}$ of medium intensity acid switches shades, it usually occurs in acid titration.