Question

1 cc of 0.1 N $HCl$ is added to 1 litre solution of sodium chloride. The $pH$of the resulting solution will be:
A.7
B.3
C.4
D.1

Answer 1

Initially, we have to calculate the normality of $HCl$ in the resulting solution. We can calculate the normality using the number of gram-equivalents of $HCl$ and volume of the solution. From the obtained normality, we can calculate the $pH$ of the solution using the concentration of ${H^ + }$.

Formula used: We can use the formula below to calculate the pH,
$pH = - \log \left[ {{H^ + }} \right]$
Here, $\left[ {{H^ + }} \right]$ is the concentration of hydrogen ions

Complete step by step answer:
Given data contains:
Volume of $HCl$is 1cc.
Normality of $HCl$ is $0.1\,N$.
Volume of sodium hydroxide is 1L.
We have to convert the volume in cc to volume in liters.
We can convert cc to liters by dividing the volume (in cc) by 1000.
$L = 1\,cc \times \dfrac{{1\,L}}{{1000\,cc}}$
$L = 0.001\,L$
The volume in litres is $0.001\,L$.
The number of gram-equivalents of $HCl$ is calculated using the normality and volume. The product of normality and volume will be the number of gram-equivalents.
$Number\,of\,gram - equivalents = Normality \times Volume$
Let us now substitute the values of normality and volume. We can calculate the number of gram-equivalents as,
$Number\,of\,gram - equivalents = Normality \times Volume$
$Number\,of\,gram - equivalents = 0.1\, \times 0.001$
$Number\,of\,gram - equivalents = 0.0001$
The number of gram-equivalents is $0.0001$.
Let us now calculate the volume of the solution. We can calculate the volume of solution by adding the volume of sodium chloride and volume of hydrochloric acid.
$Volume\,of\,solution = Volume\,of\,NaCl + Volume\,of\,HCl$
Substituting the values of volume of sodium chloride and volume of hydrochloric acid, we get,
$Volume\,of\,solution = 1\,L + 0.001\,L$
$Volume\,of\,solution = 1.001\,L$
The volume of the solution is $1.001\,L$.
Sodium chloride does not contribute to $pH$ of the solution as it is a salt.
We can now calculate the normality of the $HCl$ as,
$Normality = \dfrac{{Number\,of\,gram\,equivalents}}{{Volume\,of\,solution}}$
Substitute the values of gram equivalents and volume of solution to get the normality.
$Normality = \dfrac{{Number\,of\,gram\,equivalents}}{{Volume\,of\,solution}}$
$Normality = \dfrac{{0.0001}}{{1.001}}$
$Normality = 0.0001\,N$
$Normality = {10^{ - 4}}\,N$
The normality of the $HCl$ is ${10^{ - 4}}\,N$.
We can now calculate the $pH$ as follows,
$pH = - \log \left[ {{H^ + }} \right]$
Substitute the concentration of hydrogen ions. We get,
$pH = - \log \left[ {{H^ + }} \right]$
$pH = - \log \left[ {{{10}^{ - 4}}} \right]$
$pH = 4$
The $pH$of the solution is $4$.

So, the correct answer is Option C .

Note:
Concentration is antilogarithm of ${\text{pH}}{\text{.}}$ The concentration is calculated as,
$\left[ {{H_3}{O^ + }} \right] = {10^{ - pH}}$
For example:
Given, the $pH$ of the solution is $10.2$.
The value of $\left[ {{H_3}{O^ + }} \right]$ is calculated as,
$\left[ {{H_3}{O^ + }} \right] = {10^{ - pH}} \\\ \left[ {{H_3}{O^ + }} \right] = {10^{ - 10.2}} \\\ \left[ {{H_3}{O^ + }} \right] = 6.30 \times {10^{ - 11}}M \\\$
The $\left[ {{H_3}{O^ + }} \right]$ of the solution is $6.3 \times {10^{ - 11}}M$.