Chemistry, 11.03.2020 22:56 amberpetty4288
The value of Ka for acetic acid , CH3COOH , is 1.80×10-5 . Write the equation for the reaction that goes with this equilibrium constant. (Use H3O+ instead of H+.)
Answers: 2
![K_c=\frac{[CH_3COO^-][H_3O^+]}{[CH-3COOH][H_2O]}](/tpl/images/0543/5879/7c275.png)
![K_a=K_c\times [H_2O]=\frac{[CH_3COO^-][H_3O^+]}{[CH-3COOH]}](/tpl/images/0543/5879/125a5.png)
![[H_2O]=1](/tpl/images/0543/5879/b8579.png)
![K_a=\frac{[CH_3COO^-][H_3O^+]}{[CH-3COOH]}](/tpl/images/0543/5879/e4df0.png)
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The value of Ka for acetic acid , CH3COOH , is 1.80×10-5 . Write the equation for the reaction that...
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