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Standard XII

Chemistry

Salt of Strong Acid and Strong Bases

100 mL of 0.1...

Question

$100 m L$ of $0.1 M H C l$ is titrated against the $0.1 M N a O H$ solution. What will be the $p H$ of the solution when $100 m L N a O H$ is added ?

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Solution

The correct option is B 7
mmol of $H C l = 100 \times 0.1 = 10$
mmol of $N a O H added = 100 \times 0.1 = 10$

$H C l (a q) + N a O H (a q) \to N a C l (a q) + H_{2} O (l) I n i t i a l : 101000 F i n a l : 001010$

After titration , no acid or base is remaining as both will completely neutralizes each other and salt $N a C l$ is formed which does not hydrolyzed infact it rather hydrated in water.

So , the resultant solution will be neutral and the pH of the solution will be equal to 7.

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Similar questions

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Salt of Strong Acid and Strong Bases

Standard XII Chemistry

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$100 m L$ of $0.1 M H C l$ is titrated against the $0.1 M N a O H$ solution. What will be the $p H$ of the solution when $100 m L N a O H$ is added ?