So let's say we have $200$ mL of $1$M $CH_3COOH$ solution. In this solution we have the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$. To that we add $100$mL of $1$M $NaOH$ solution.Then after reacting we get a buffer solution where $[CH_3COOH] = [CH_3COO^-] = \frac{1}{3} moldm^{-3}$. In moles, we have $0.1$ moles of both $CH_3COOH$ and $CH_3COO^-$.

Then suppose we add $x$ mL of $0.1$M NaOH. This is, in effect, reacting $0.1$ moles of $CH_3COOH$ with $0.0001x$ moles of $NaOH$, the result of which is that now $n(CH_3COOH) = 0.1 - 0.0001x$ and $n(CH_3COO^-) = 0.1 + 0.0001x$.

Could we not then model, by the Henderson-Hasselbach equation, $pH = pKa + log(\frac{0.1 +0.0001x}{0.1 - 0.0001x}) = 4.76 +log(\frac{0.1 +0.0001x}{0.1 - 0.0001x})$?

But this would mean that for the pH of the buffer solution to rise by $1$, we would need $818.82 cm^3$ of $0.1$M NaOH solution, which sounds absurd. Moreover, for different acids, like propanoic acid and butanoic acid, the same line of logic could be used to deduce that the volume required to increase the pH by 1 would be the same for all, but they have different buffering capacities. So what's wrong with the logic?

So let's say we have $200\ \mathrm{mL}$ of $1\ \mathrm M$ $\ce{CH3COOH}$ solution. In this solution we have the equilibrium $\ce{CH3COOH <=> CH3COO- + H+}$. To that we add $100\ \mathrm{mL}$ of $1\ \mathrm M$ $\ce{NaOH}$ solution. Then after reacting we get a buffer solution where $[\ce{CH3COOH}] = [\ce{CH3COO-}] = \frac{1}{3}\ \mathrm{mol\ dm^{-3}}$. In moles, we have $0.1\ \mathrm{mol}$ of both $\ce{CH3COOH}$ and $\ce{CH3COO-}$.

Then suppose we add $x\ \mathrm{mL}$ of $0.1\ \mathrm M$ $\ce{NaOH}$. This is, in effect, reacting $0.1\ \mathrm{mol}$ of $\ce{CH3COOH}$ with $0.0001 x\ \mathrm{mol}$ of $\ce{NaOH}$, the result of which is that now $n(\ce{CH3COOH}) = (0.1 - 0.0001 x)\ \mathrm{mol}$ and $n(\ce{CH3COO-}) = (0.1 + 0.0001 x)\ \mathrm{mol}$.

Could we not then model, by the Henderson-Hasselbalch equation, $\mathrm{pH} = \mathrm pK_\mathrm a + \log\left(\dfrac{0.1 + 0.0001 x}{0.1 - 0.0001 x}\right) = 4.76 + \log\left(\dfrac{0.1 + 0.0001 x}{0.1 - 0.0001 x}\right)$?

But this would mean that for the pH of the buffer solution to rise by $1$, we would need $818.82\ \mathrm{cm^3}$ of $0.1\ \mathrm M$ $\ce{NaOH}$ solution, which sounds absurd. Moreover, for different acids, like propanoic acid and butanoic acid, the same line of logic could be used to deduce that the volume required to increase the $\mathrm{pH}$ by $1$ would be the same for all, but they have different buffering capacities. So what's wrong with the logic?

So let's say we have $200$ mL of $1$M $CH_3COOH$ solution. In this solution we have the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$. To that we add $100$mL of $1$M $NaOH$ solution.Then after reacting we get a buffer solution where $[CH_3COOH] = [CH_3COO^-] = \frac{1}{3} moldm^{-3}$. In moles, we have $0.1$ moles of both $CH_3COOH$ and $CH_3COO^-$.

Then suppose we add $x$ mL of $0.1$M NaOH. This is, in effect, reacting $0.1$ moles of $CH_3COOH$ with $0.0001x$ moles of $NaOH$, the result of which is that now $n(CH_3COOH) = 0.1 - 0.0001x$ and $n(CH_3COO^-) = 0.1 + 0.0001x$.

Could we not then model, by the Henderson-Hasselbach equation, $pH = pKa + log(\frac{0.1 +0.0001x}{0.1 - 0.0001x}) = 4.76 +log(\frac{0.1 +0.0001x}{0.1 - 0.0001x})$?

But this would mean that for the pH of the buffer solution to rise by $1$, we would need $818.82 cm^3$ of $0.1$M NaOH solution, which sounds absurd.

So let's say we have $200$ mL of $1$M $CH_3COOH$ solution. In this solution we have the equilibrium $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$. To that we add $100$mL of $1$M $NaOH$ solution.Then after reacting we get a buffer solution where $[CH_3COOH] = [CH_3COO^-] = \frac{1}{3} moldm^{-3}$. In moles, we have $0.1$ moles of both $CH_3COOH$ and $CH_3COO^-$.

Then suppose we add $x$ mL of $0.1$M NaOH. This is, in effect, reacting $0.1$ moles of $CH_3COOH$ with $0.0001x$ moles of $NaOH$, the result of which is that now $n(CH_3COOH) = 0.1 - 0.0001x$ and $n(CH_3COO^-) = 0.1 + 0.0001x$.

Could we not then model, by the Henderson-Hasselbach equation, $pH = pKa + log(\frac{0.1 +0.0001x}{0.1 - 0.0001x}) = 4.76 +log(\frac{0.1 +0.0001x}{0.1 - 0.0001x})$?

But this would mean that for the pH of the buffer solution to rise by $1$, we would need $818.82 cm^3$ of $0.1$M NaOH solution, which sounds absurd. Moreover, for different acids, like propanoic acid and butanoic acid, the same line of logic could be used to deduce that the volume required to increase the pH by 1 would be the same for all, but they have different buffering capacities. So what's wrong with the logic?