Given Ka values of 5.76×10^−10 and 4.8×10^−10 for NH+4 and HCN respectively. What is the equilibrium constant for the following reaction? NH+4(aq.)+CN−(aq.)⇔NH3(aq.)+HCN(aq.)

Sahay Sir > Question Answers > Given Ka values of 5.76×10^−10 and 4.8×10^−10 for NH+4 and HCN respectively. What is the equilibrium constant for the following reaction? NH+4(aq.)+CN−(aq.)⇔NH3(aq.)+HCN(aq.)
Given Ka values of 5.76×10^−10 and 4.8×10^−10 for NH+4 and HCN respectively. What is the equilibrium constant for the following reaction? NH+4(aq.)+CN−(aq.)⇔NH3(aq.)+HCN(aq.)
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Given Ka values of 5.76×10^−10 and 4.8×10^−10 for NH+4 and HCN respectively. What is the equilibrium constant for the following reaction? NH+4(aq.)+CN−(aq.)⇔NH3(aq.)+HCN(aq.) Given Ka values of 5.76×10^−10 and 4.8×10^−10 for NH+4 and HCN respectively. What is the equilibrium constant for the following reaction? NH+4(aq.)+CN−(aq.)⇔NH3(aq.)+HCN(aq.) November 12, 2020 Category: Chapter 6 - Ionic Equilibrium (Level 1 [...]
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Tags: Given Ka values of 5.76×10^−10 and 4.8×10^−10 for NH+4 and HCN respectively. What is the equilibrium constant for the following reaction? NH+4(aq.)+CN−(aq.)⇔NH3(aq.)+HCN(aq.) ,