
1 c.c of 0.1 N HCl is added to a 1 liter solution of sodium chloride. The pH of the resulting solution will be:
A) 7
B) 0
C) 10
D) 4
Answer
512.1k+ views
Hint: The pH of any solution can be determined from the concentration of \[{{H}^{+}}\] in the solution. The concentration of \[{{H}^{+}}\]will be determined from the concentration of HCl. There is a formula to calculate the pH of the solution.
pH = -log[\[{{H}^{+}}\]]
The units to express the concentration of \[{{H}^{+}}\] ion is \[mol.\text{ }{{L}^{-1}}\] or M.
Complete answer:
In the question it is given that volume of HCL = 1 c.c = 0.001 L
Volume of NaCl = 1 L
Normality of HCL = 0.1 N
We have to find the pH of the resulting solution.
Normality volume = (0.1) (0.001)
= 0.0001 L = \[{{10}^{-4}}\]L
Volume of total solution = Volume of NaCl + Volume of HCL
= 1 + 0.001 = 1.001 L
Meanwhile sodium chloride is a salt and is neutral in nature. So, it does not affect the pH of a solution.
Normality of HCL in a resulting solution = \[\dfrac{0.0001}{1.001}\]
= 0.0001 N
pH = -log[\[{{H}^{+}}\]]
= - log (0.0001)
= - log (\[{{10}^{-4}}\])
= 4
Therefore the pH of the solution is 4.
So, the correct option is D.
Note: If an acid solution contains sodium chloride, the concentration of the acid solution is not going to change because sodium chloride is a salt and is neutral in nature. Due to its neutral behavior it does not affect the concentration of the acid solution and its presence is negligible.
pH = -log[\[{{H}^{+}}\]]
The units to express the concentration of \[{{H}^{+}}\] ion is \[mol.\text{ }{{L}^{-1}}\] or M.
Complete answer:
In the question it is given that volume of HCL = 1 c.c = 0.001 L
Volume of NaCl = 1 L
Normality of HCL = 0.1 N
We have to find the pH of the resulting solution.
Normality volume = (0.1) (0.001)
= 0.0001 L = \[{{10}^{-4}}\]L
Volume of total solution = Volume of NaCl + Volume of HCL
= 1 + 0.001 = 1.001 L
Meanwhile sodium chloride is a salt and is neutral in nature. So, it does not affect the pH of a solution.
Normality of HCL in a resulting solution = \[\dfrac{0.0001}{1.001}\]
= 0.0001 N
pH = -log[\[{{H}^{+}}\]]
= - log (0.0001)
= - log (\[{{10}^{-4}}\])
= 4
Therefore the pH of the solution is 4.
So, the correct option is D.
Note: If an acid solution contains sodium chloride, the concentration of the acid solution is not going to change because sodium chloride is a salt and is neutral in nature. Due to its neutral behavior it does not affect the concentration of the acid solution and its presence is negligible.
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