Question

Following solutions were prepared by mixing different volume of ${\text{NaOH}}$ and ${\text{HCl}}$ of different concentration:
A.$60{\text{ mL}}\dfrac{{\text{M}}}{{10}}{\text{HCl}} + 40\,{\text{mL}}\dfrac{{\text{M}}}{{10}}{\text{NaOH}}$
B.${\text{55 mL}}\dfrac{{\text{M}}}{{10}}{\text{HCl}} + 45{\text{mL}}\dfrac{{\text{M}}}{{10}}{\text{NaOH}}$
C.${\text{75 mL}}\dfrac{{\text{M}}}{5}{\text{HCl}} + 25\,{\text{mL}}\dfrac{{\text{M}}}{5}{\text{NaOH}}$
D.${\text{100 mL}}\dfrac{{\text{M}}}{{10}}{\text{HCl}} + 100\,{\text{mL}}\dfrac{{\text{M}}}{{10}}{\text{NaOH}}$
pH of which one of them will be equal to 1?

Answer 1

First of all we need to calculate the number of moles or millimole of ${\text{NaOH}}$and ${\text{HCl}}$and neutralize them. Then using the net amount of acid and base formed in the total volume we will get the concentration. Using the concentration in pH formula we will get pH.
Formula used: ${\text{pH}} = - {\text{log}}[{{\text{H}}^ + }]$
${\text{number of millimole}} = {\text{molarity}} \times {\text{volume}}({\text{mL}})$

Complete step by step answer:
Let us first consider the option A, $60{\text{ mL}}\dfrac{{\text{M}}}{{10}}{\text{HCl}} + 40\,{\text{mL}}\dfrac{{\text{M}}}{{10}}{\text{NaOH}}$
The number of millimoles of ${\text{HCl}}$ will be,
${\text{number of millimoles of HCl}} = \dfrac{1}{{10}} \times 60 = 6$
The number of millimoles of ${\text{NaOH}}$ will be,
${\text{number of millimoles of NaOH}} = \dfrac{1}{{10}} \times 40 = 4$
6 millimole of ${\text{HCl}}$ will neutralize 4 millimole of ${\text{NaOH}}$ and hence the resulting solution formed will contain 2 millimoles of ${\text{HCl}}$.
Now the total volume has become 100 mL. So the concentration of ${{\text{H}}^ + }$is:
${\text{[}}{{\text{H}}^ + }] = \dfrac{{2{\text{ millimole}}}}{{100{\text{ milliliter}}}} = 0.02$
We will calculate pH as:
${\text{pH}} = - {\text{log}}[0.02] = 1.69$
In option B, ${\text{55 mL}}\dfrac{{\text{M}}}{{10}}{\text{HCl}} + 45{\text{mL}}\dfrac{{\text{M}}}{{10}}{\text{NaOH}}$
The number of millimoles of ${\text{HCl}}$ will be,
${\text{number of millimoles of HCl}} = \dfrac{1}{{10}} \times 55 = 5.5$
The number of millimoles of ${\text{NaOH}}$ will be,
${\text{number of millimoles of NaOH}} = \dfrac{1}{{10}} \times 45 = 4.5$
$5.5$ millimole of ${\text{HCl}}$ will neutralize $4.5$ millimole of ${\text{NaOH}}$ and hence the resulting solution formed will contain 1 millimoles of ${\text{HCl}}$.
Now the total volume has become 100 mL. So the concentration of ${{\text{H}}^ + }$is:
${\text{[}}{{\text{H}}^ + }] = \dfrac{{{\text{1 millimole}}}}{{100{\text{ milliliter}}}} = 0.01$
We will calculate pH as:
${\text{pH}} = - {\text{log}}[0.01] = 2$
In option C, ${\text{75 mL}}\dfrac{{\text{M}}}{5}{\text{HCl}} + 25\,{\text{mL}}\dfrac{{\text{M}}}{5}{\text{NaOH}}$
The number of millimoles of ${\text{HCl}}$ will be,
${\text{number of millimoles of HCl}} = \dfrac{1}{5} \times 75 = 15$
The number of millimoles of ${\text{NaOH}}$ will be,
${\text{number of millimoles of NaOH}} = \dfrac{1}{5} \times 25 = 5$
15 millimolar of ${\text{HCl}}$ will neutralize 5 millimole of ${\text{NaOH}}$ and hence the resulting solution formed will contain 10 millimoles of ${\text{HCl}}$.
Now the total volume has become 100 mL. So the concentration of ${{\text{H}}^ + }$is:
${\text{[}}{{\text{H}}^ + }] = \dfrac{{{\text{10 millimolar}}}}{{100{\text{ milliliter}}}} = 0.1$
We will calculate pH as:
${\text{pH}} = - {\text{log}}[0.1] = 1$

Hence the correct option is C.

Note:
M given in the option is a unit of concentration that is molarity and not a variable. In the fourth option equal volume of both strong acid and base are taken with the same molarity. Both acid and base will neutralize each other and hence the pH of the solution will be neutral.