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What is the pH of dil. HCl solution with concentration ${{10} ^ {-8}} $mol/L?
(a) 7
(b) 8
(c) 6.98
(d) 10

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Last updated date: 13th Jun 2024
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Answer
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Hint: Apply the formula of pH to calculate the pH of acid HCl and keep in mind in dilute acid cases, the concentration of hydrogen ions of water is also considered. Now, solve it.

Complete answer: Hydrochloric acid is a colorless inorganic chemical substance with the chemical formula as HCl. It is a strong acid band that is completely dissociated (breaks down) in water.
Be before calculation, pH we should know what it is: pH 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.
In aqueous solution (which consists water as solvent)of a strong acid:
Acid is present as the solute and water is present in the solvent form.
An aqueous solution of strong acid can be made into the dilute solution by the addition of more solvent i.e., water to the solution.
When an aqueous solution of strong acid is diluted by adding water then the concentration of $H{}_{a}^{+}$ in the solution decreases and thus, the pH of the solution increases.

 The pH of the water is always 7 and is considered only when:
When the concentration of $H{}_{{}}^{+}$ (from acid) is greater than the water, so in that case we can neglect the $H{}_{{}}^{+}$ (from water)i.e. ${{10}^{-7}}$ value of water.
 But when the concentration of $H{}_{{}}^{+}$ (from acid) is lesser than the water, so in that case we cannot neglect the $H{}_{{}}^{+}$ (from water)i.e. ${{10}^{-7}}$ value of water is to be taken ( in case of dilute solutions).
Now, we can see that in the statement the concentration of $H{}_{{}}^{+}$ from HCl is less because the solution is dilute in which the water is in present in excess so, we cannot neglect the $H{}_{{}}^{+}$ (from water) i.e. ${{10}^{-7}}$ value of water

In ${{10} ^ {-8}} $M HCl solution (dilute) given in the statement:

Total [H+] =${{10} ^ {-8}} $ (from acid) +${{10} ^ {-7}} $ (from water)

                   =1.1×${{10} ^ {-7}} $
Now, applying the formula of pH, we get:

              pH= − log [$H {} _ {{}} ^ {+} $]

                   = −log (1.1×${{10} ^ {-7}} $)
                   = − log 1.1 +7 log10
Since log10=1
                    = − log 1.1+ 7
From the logarithm tables, we get to know that the log1.1= 0.10414, then:
                    = − 0.10414 + 7
                    = 6.89586
                    ≈6.98(approx.)

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.