84. pH of a neutral solution at 360 K is (K at 360 K is
3.6 10-13)
Answers
Answer
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Explanation:
We know,at equilibrium ,
Kc = [SO3]2[SO2]2[O2] = 100
As moles of SO2 is equal to that of SO3 ,
[SO2] = [SO3]
[O2] = 1100moles of SO3moles of SO2 = 2We again have, Kc = [SO3]2[SO2]2 [O2] = 100 (moles of SO3moles of SO2)2(10moles of O2) = 100 22 × 10moles of O2 = 100 mole of O2 = 0.4
Mole of O210 = 1100 (vol = 10 L) Mole of O2 = 0.1
2. Further,if moles of SO3 are twice that of SO2,
The Ionic product of water is defined as below, It is the product of concentration of Hydronium ion and Hydroxy ion.
Kw = [H30+][OH-]
For a neutral solution
[H30+]= [OH-]
Let the concentration of Hydronium ion is x.
So, Ionic product of water will be - Kw = [H30+][OH-]
= Kw = x ²
= x² = 3.6 × 10^ - 13
= x² = 36 × 10^ -14
= x = 6 × 10 ^ - 7
Thus, the concentration of Hydronium ion - [H30+][OH-] = 6 × 10`-7
ph = -log ( 6 × 10`-7)
ph = -log 6 + log 10`-7
ph = - ( -7 + log 6)
ph = 7 - log (6)
Therefore, the pH of a neutral solution at 360K is 7 - log6,