The equilibrium constant for the ionization of RNH2 (g) in water as RNH2(g)+H2O(l)⇔RNH+3(aq)+OH−(aq) is 8×10^−6 at 25^∘C. find the pH of a solution at equilibrium when pressure of RNH2(g) is 0.5 bar :

Sahay Sir > Question Answers > The equilibrium constant for the ionization of RNH2 (g) in water as RNH2(g)+H2O(l)⇔RNH+3(aq)+OH−(aq) is 8×10^−6 at 25^∘C. find the pH of a solution at equilibrium when pressure of RNH2(g) is 0.5 bar :
The equilibrium constant for the ionization of RNH2 (g) in water as RNH2(g)+H2O(l)⇔RNH+3(aq)+OH−(aq) is 8×10^−6 at 25^∘C. find the pH of a solution at equilibrium when pressure of RNH2(g) is 0.5 bar :
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Nov
The equilibrium constant for the ionization of RNH2 (g) in water as RNH2(g)+H2O(l)⇔RNH+3(aq)+OH−(aq) is 8×10^−6 at 25^∘C. find the pH of a solution at equilibrium when pressure of RNH2(g) is 0.5 bar : The equilibrium constant for the ionization of RNH2 (g) in water as RNH2(g)+H2O(l)⇔RNH+3(aq)+OH−(aq) is 8×10^−6 at 25^∘C. find the pH of a [...]
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Tags: The equilibrium constant for the ionization of RNH2 (g) in water as RNH2(g)+H2O(l)⇔RNH+3(aq)+OH−(aq) is 8×10^−6 at 25^∘C. find the pH of a solution at equilibrium when pressure of RNH2(g) is 0.5 bar : ,