A solution has been prepared by dissolving 0.63g of nitric acid in 100ml. What is its pH value? Assume that the acid is completely dissociated.
Answers
Answered by
49
The concentration of nitric acid HNO₃ solution
= 0.63g in 100 ml
= 6.3g per litre
= 0.1 moles per litre [ As mol. wt of HNO₃ = 63 ]
= 10⁻¹ M
Now, HNO₃ completely ionizes as
HNO₃ + H₂O ===>> H₃O⁺ + NO₃⁻
[ H₃O⁺ ] = [ HNO₃ ] = 10⁻¹ M
So, pH = - log [ H₃O⁺ ]
pH = - log [ 10⁻¹ ]
pH = 1
= 0.63g in 100 ml
= 6.3g per litre
= 0.1 moles per litre [ As mol. wt of HNO₃ = 63 ]
= 10⁻¹ M
Now, HNO₃ completely ionizes as
HNO₃ + H₂O ===>> H₃O⁺ + NO₃⁻
[ H₃O⁺ ] = [ HNO₃ ] = 10⁻¹ M
So, pH = - log [ H₃O⁺ ]
pH = - log [ 10⁻¹ ]
pH = 1
Answered by
7
Answer:
Explanation:
See
the molar mass of HNO3 = 63 g/mole
So 0.063 g is 0.063/63 = 10^-3 moles in 1 liter = 10^-3 M.
And the H+ concentration is also 10^-3 M because HNO3 is a strong acid.
In water [H+]*[OH-] = 10^-14 M^2 = Kw. So [OH- ]= 10^-14 / [H+] = 10^-14 / 10^-3 = 10^-11 M
therefore
pH= - log[H+]
=3
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