# Why there is an increase in the pH at the starting of a titration between a weak acid and a strong base?

Why there is increase in the pH at the starting of a titration between a weak acid and a strong base even though there is acid already present to neutralize it? My intuition says that there should not be even a slight increase in pH until all of the weak acid is dissociated and neutralized by the base • The StackExchange network expects that you have thoroughly searched and thought about the topic and have provided explicit compact summary of partial answers/ideas/thoughts you have got until now. That will prevent responders to tell you what you already know or what can be easily found. Effort not shown may be considered as effort not done and such a question may get closed. How do I ask a good question?. Jan 22 at 12:05
• What does the equation pH = pKa + log([A-]/[HA]) say to you? Jan 22 at 12:13
• @Poutnik . more acid will dissociate as pH increases. Jan 22 at 12:35
• @NikhilVerma It has also the other side - increasing acid dissociation ratio by acid neutralization increases pH, Jan 22 at 14:55

Let's start from the basic formula : $$\mathrm{p}H = \mathrm{p}K_a + \log\mathrm{\frac{[A^-]}{[HA]} = \mathrm{p} K_a + {log{\frac{\mathrm{[A_o]} + n}{[HA]_o - n}}}}$$ The derivate of log$$x$$ is $$1/x$$. So when introducing $$n$$ mole $$\ce{NaOH}$$, we get a curve with a slope equal to : $$\frac{d\mathrm pH}{dn} = \mathrm{\frac{[HA]}{[A^-]}} = \frac{[\mathrm{H^+]}}{K_{a}}$$
At the beginning of the titration, when $$n$$ is small, the acidic concentration $$\ce{[H^+]}$$ is high. So the slope of the curve is steep. When $$n$$ grows bigger, $$\ce{[H+]}$$ becomes smaller. As a consequence, the slope of the curve decreases.