3 My main concern was the correction of Mr Hasselba**l**ch, I also cleaned the maths markup while I was at it
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  1. In general, acid + $\ce{H_2O} \rightleftharpoons$ base +: $\ce{H_3O^+}$$\ce{acid + H2O <=> base + H_3O^+}$
  2. It is called the Henderson-HasselbachHasselbalch equation $pH = pK_a + \log\frac{[base]}{[acid]}$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$
  3. I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_aK_b\frac{1}{K_w}$$K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$
  1. In general, acid + $\ce{H_2O} \rightleftharpoons$ base + $\ce{H_3O^+}$
  2. It is called the Henderson-Hasselbach equation $pH = pK_a + \log\frac{[base]}{[acid]}$
  3. I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_aK_b\frac{1}{K_w}$
  1. In general: $\ce{acid + H2O <=> base + H_3O^+}$
  2. It is called the Henderson-Hasselbalch equation $\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$
  3. I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$
2 formatting
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  1. In general, acid + $H_2O \rightleftharpoons$$\ce{H_2O} \rightleftharpoons$ base + $H_3O^+$$\ce{H_3O^+}$
  2. It is called the hendersonHenderson-hasselbachHasselbach equation $pH = pK_a + log\frac{[base]}{[acid]}$$pH = pK_a + \log\frac{[base]}{[acid]}$
  3. I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_aK_b\frac{1}{K_w}$
  1. In general, acid + $H_2O \rightleftharpoons$ base + $H_3O^+$
  2. It is called the henderson-hasselbach equation $pH = pK_a + log\frac{[base]}{[acid]}$
  3. I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_aK_b\frac{1}{K_w}$
  1. In general, acid + $\ce{H_2O} \rightleftharpoons$ base + $\ce{H_3O^+}$
  2. It is called the Henderson-Hasselbach equation $pH = pK_a + \log\frac{[base]}{[acid]}$
  3. I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_aK_b\frac{1}{K_w}$
1
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  1. In general, acid + $H_2O \rightleftharpoons$ base + $H_3O^+$
  2. It is called the henderson-hasselbach equation $pH = pK_a + log\frac{[base]}{[acid]}$
  3. I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_aK_b\frac{1}{K_w}$