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# The pH of the solution obtained on neutralisation of 40ml of 0.1 NaOH with 40ml of 0.1M $C{{H}_ {3}} COOH$ is:(A) 7(B) 8(C) 6(D) 3

Last updated date: 20th Jun 2024
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Hint: For determining pH: We know that the outcome of a pH-measurement is determined by a consideration between the number of ${{H}^ {+}}$ ions and the number of hydroxide $O{{H}^ {-}}$ ions. When the number of ${{H}^ {+}}$ions equals the number of $O{{H}^ {-}}$ ions, the water is neutral.

Complete step by step solution:
We have been provided with a strong base NaOH and a weak acid $C{{H}_ {3}} COOH$,
We need to find the pH of the solution obtained on neutralisation of 40ml of 0.1 NaOH with 40ml of 0.1M $C{{H}_ {3}} COOH$,
So, for that:
We know that acid and base react to form salt and water, and the reaction is called neutralisation reaction,
Neutralisation is a chemical reaction in which acid and a base react quantitatively with each other. In a reaction in water, neutralization results in there being no excess of hydrogen or hydroxide ions present in the solution.
But in this case the base is strong and acid is weak,
So, the effect of base would be more than acid,
So, the solution formed in neutralisation of 40ml of 0.1 NaOH with 40ml of 0.1M $C{{H}_ {3}} COOH$would be basic in nature,
So, we know that the basic solution has a pH more than 7,
So, we can say that the pH of the solution obtained on neutralisation of 40ml of 0.1 NaOH with 40ml of 0.1M $C{{H}_ {3}} COOH$ is 8.

Therefore, we can conclude that option (B) is correct.

Note: The Ph of neutral solution is 7, and that of acidic solution is less than 7, It is common that the pH scale is argued to range from 0-14 or perhaps 1-14, but neither is correct. The pH range does not have an upper or lower bound.