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 3 My main concern was the correction of Mr Hasselba**l**ch, I also cleaned the maths markup while I was at it edited May 26 '15 at 9:43 Martin - マーチン♦ 34.7k99 gold badges117117 silver badges245245 bronze badges In general, acid + $$\ce{H_2O} \rightleftharpoons$$ base +: $$\ce{H_3O^+}$$$$\ce{acid + H2O <=> base + H_3O^+}$$ It is called the Henderson-HasselbachHasselbalch equation $$pH = pK_a + \log\frac{[base]}{[acid]}$$$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$$ I believe the equilibrium constant for a such a neutralisation would look like $$K_n = K_aK_b\frac{1}{K_w}$$$$K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$$ In general, acid + $$\ce{H_2O} \rightleftharpoons$$ base + $$\ce{H_3O^+}$$ It is called the Henderson-Hasselbach equation $$pH = pK_a + \log\frac{[base]}{[acid]}$$ I believe the equilibrium constant for a such a neutralisation would look like $$K_n = K_aK_b\frac{1}{K_w}$$ In general: $$\ce{acid + H2O <=> base + H_3O^+}$$ It is called the Henderson-Hasselbalch equation $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$$ I believe the equilibrium constant for a such a neutralisation would look like $$K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$$ 2 formatting edit approved May 24 '15 at 6:43 user15489 In general, acid + $$H_2O \rightleftharpoons$$$$\ce{H_2O} \rightleftharpoons$$ base + $$H_3O^+$$$$\ce{H_3O^+}$$ It is called the hendersonHenderson-hasselbachHasselbach equation $$pH = pK_a + log\frac{[base]}{[acid]}$$$$pH = pK_a + \log\frac{[base]}{[acid]}$$ I believe the equilibrium constant for a such a neutralisation would look like $$K_n = K_aK_b\frac{1}{K_w}$$ In general, acid + $$H_2O \rightleftharpoons$$ base + $$H_3O^+$$ It is called the henderson-hasselbach equation $$pH = pK_a + log\frac{[base]}{[acid]}$$ I believe the equilibrium constant for a such a neutralisation would look like $$K_n = K_aK_b\frac{1}{K_w}$$ In general, acid + $$\ce{H_2O} \rightleftharpoons$$ base + $$\ce{H_3O^+}$$ It is called the Henderson-Hasselbach equation $$pH = pK_a + \log\frac{[base]}{[acid]}$$ I believe the equilibrium constant for a such a neutralisation would look like $$K_n = K_aK_b\frac{1}{K_w}$$ 1 answered May 24 '15 at 6:38 getafix 6,99333 gold badges2424 silver badges5656 bronze badges In general, acid + $$H_2O \rightleftharpoons$$ base + $$H_3O^+$$ It is called the henderson-hasselbach equation $$pH = pK_a + log\frac{[base]}{[acid]}$$ I believe the equilibrium constant for a such a neutralisation would look like $$K_n = K_aK_b\frac{1}{K_w}$$