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Question: Which of the following plot represents the graph of pH against volume of alkali added in the titrati...

Which of the following plot represents the graph of pH against volume of alkali added in the titration of NaOHNaOH and HClHCl
A.
B.
C.
D.

Explanation

Solution

We need to know that pHpH is mainly used to find out the solution which is acidic, basic or neutral. And the range of pHpH from one to 1414 . The solution is neutral if the pHpH is 77 . And if it is less than 77 means, the solution is acidic and pHpH is greater than seven means, the solution is basic. And pHpH mainly measures the relative amount of free hydrogen which is present in the water.

Complete answer:
While adding alkali to acid, there occurs a change in pHpH. And the acidity of acid becomes decreasing and the basicity will start to increase. But there is no sudden increase in the pHpH Hence, option (A) is incorrect.
Here, the base is added to the acid solution. Hence, pHpHof the solution is low at the initial stage. But, in this graph, the plot of pHpH starts from its higher range. Hence, option (B) is correct.
When sodium hydroxide is added to hydrochloric acid, pHpH of the solution remains the same at the initial stage. After that it suddenly increases pHpH. Then it became almost the same. After that there is no change in the pHpH. Therefore, the graph of pH against volume of alkali added in the titration of NaOHNaOH and HClHCl is,

Hence, option (C) is correct.
This is not the correct graphical representation of pH against volume. Hence, the option (D) is incorrect.

Hence, option (C) is correct.

Note:
We need to know that when the alkali and acid are mixing in the right amounts, the reaction is wound up with the formation of a neutral solution. If water or base is added to an acid, the pH of the acid will be increasing. Because, there is a decrease in the acidity and the acid becomes less acidic. If the water is added to an alkali solution, the pH of the base becomes near 77 due to the decrease in the concentration of hydroxide ions.