Question

Given, 1cc of 0.1 N HCl is added to the 999 cc solution of NaCl. The pH of the resulting solution will be:
(a) 7
(b) 3
(c) 4
(d) 1

Answer 1

Hint: First , of all we will have to calculate the volume of solution and after that, we will calculate the normality of HCl in that solution by applying the formula as: $Normality=equivalent\text{ }weight\times volume$ and then we can easily find the pH through this and NaCl will not contribute towards it as it is a salt. Now, solve it.

Complete answer:
First of all, let’s discuss what is pH. By the term pH we mean that it gives us the measure of acid/base strength of any solution. pH scale ranges from 0-14, 7 is neutral, below 7 it represents acidic and above 7 it represents basic.
Now considering the statement;
Volume of the HCl solution= 1cc ( given)
$\begin{align}
& =1\text{ c}{{\text{m}}^{3}} \\
& =0.001\text{ L ( 1L=1000 c}{{\text{m}}^{3}}) \\
\end{align}$
Normality of the HCl = 0.1 N
Now, as we know that;
$Normality=equivalent\text{ }weight\times volume$ ------------(1)
From this we can easily calculate the equivalent weight of the HCl as;
$\begin{align}
  & equivalent\text{ }weight=\dfrac{volume}{normality} \\
 & \text{ =}\dfrac{0.1}{0.001} \\
 & \text{ = 0}\text{.0001} \\
\end{align}$
Now, we know that Volume of NaCl solution= 999c (given)
$\begin{align}
  & =999\text{ c}{{\text{m}}^{3}} \\
 & =0.999\text{ L ( 1L=1000 c}{{\text{m}}^{3}}) \\
 & =1\text{ L} \\
\end{align}$
So, the total volume of the solution is ;
$\begin{align}
  & \text{ Total volume}=\text{ volume of NaCl + volume of HCl} \\
 & \text{ = 1+ 0}\text{.001 L} \\
 & \text{ = 1}\text{.001 L} \\
\end{align}$
Now, we will calculate the normality of HCl solution in 1.001 L by using the normality formula from equation(1) , we get;
$\begin{align}
  & Normality=0.0001\times 1.001 \\
 & \text{ = 0}\text{.0001 }N \\
\end{align}$
Now, we will calculate the pH of the solution by using the formula as;
$pH=-\log [{{H}^{+}}]$
Since, NaCl is a salt and so, it will not be added while calculating the pH of the solution and only HCl will contribute towards it.
Then, pH is;
$\begin{align}
  & pH=-\log [{{H}^{+}}] \\
 & \text{ = -log(0}\text{.0001)} \\
 & \text{ =-log 1}{{\text{0}}^{-4}}+\text{log1} \\
 & \text{ = -(-4)+0 ( log 10=1 ,log 1=0)} \\
 & \text{ =4} \\
\end{align}$
Hence, 1cc of 0.1 N HCl when added to the 999 cc solution of NaCl ,the pH of the resulting solution is 4.

So, option (c) is correct.

Note:
The pH of acid must always be less than 7 and in case, the acid is more diluted then the concentration of hydrogen ion of the water molecules is also considered while calculating the pH.