1.

Amol took 10 mL of 2.2 times 10-5 M hydrochloric acid solution. He then diluted it to 1 litre. He found that the pH of diluted solution is a) 4.7b) 6.7 c) 4.5 d) 6.5

Answer»

Answer is : d) 6.5

Correct option is (D) 6.5

using molarity equation-

M1V2 = M2V2

10 ml x 2.2 x 10-5 M = 1000 mL x M2

M2 = 2.2 x 10-7M of HCl

\(\therefore\) final concentration of HCl = 2.2 x 10-7 M.

Here, we also consider the H+ions come from water dessociation.

We know that

[H+] = [\(\bar O\)H] = 1 x 10-7

\(\therefore\) Total H+ concentration in solution

 = (2.2 x 10-7) + (1 x 10-7) M

 = 3.2 x 10-7 M

\(\therefore\) PH of solution = - log[H+]

 = -[log(3.2) + log 10-7]

 = -0.5 + 7

 = 6.5

Hence, the PH of final solution will be 6.5



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