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10mL of 0.1NHCl is added to 990mLsolution ofNaCl. The pHof the resulting solution is – (A) zero (B) 3 (C) 7 (D) 10

10mL of 0.1NHCl is added to 990mLsolution ofNaCl. The pHof the resulting solution is –
(A) zero (B) 3 (C) 7 (D) 10

Grade:10

1 Answers

Arun
25750 Points
5 years ago
Volume of HCL = 10 cm³ = 10/1000 = 0.01 L
Volume of NaCl = 990 cm³ = 99/01000 = 0.99 L
Normality of HCL = 0.1 N
pH of the resulting solution = ?
Solution:
Normality volume = (0.1) (0.01)
Normality volume = 0.001 L
Volume of solution = Volume of NaCl + Volume of HCL
Volume of solution = 1 + 0.01
Volume of solution = 1.01 L
Since NaCl is a salt and is neutral so it does not effect pH of a solution.
Now,
Normality of HCL in a resulting solution = 0.001 / 1.01
Normality of HCL in a resulting solution = 0.01 N
So,
pH = - log [H⁺]
pH = - log (0.001)
pH = 3
which shows that the pH of the resulting solution is e

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