# Why is the gradient of the curve of a strong base titrated with strong acid small up until equivalence?

I understand the cause of the sharp decrease in pH upon reaching the equivalence point, but I don't understand why addition of relatively large amounts of the strong acid would be accompanied with small changes in pH before that.

When looking at how pH changes when adding a strong acid to water there is a very fast drop in pH, as to be expected due to the logarithmic nature of the pH scale. The pH scale (I believe) also explains the flat region at the start of the curve for titrating a strong acid with a strong base.

But I don't understand it for the titration in question. I presume it must be something to do with the neutralisation but I'm not sure.

• Below pH=7, you have H+ ions because you physically added them. Above pH=7, you have H+ ions because of the autoprotolysis of water. – Karl Dec 14 '20 at 9:22

Let's calculate the $$p$$H by changing the volume of $$\ce{HCl}$$ added. I will use the unit millimole ($$1 mmol = 0.001 mol$$), which is more practical than the mole.

In the very beginning the solution contains $$50 mL~ \ce{NaOH}~ 0.1 M$$, or $$5 mmol$$ $$\ce{NaOH}$$. Let's take, for example, the point when you have added $$30 mL~ \ce{HCl} ~0.1 M$$ (or $$3.0 mmol ˙\ce{HCl}$$). After this addition, the solution contains $$5 - 3.0 = 2.0 mmol$$ NaOH in a volume $$50 mL + 30 mL = 80 mL$$. The concentration of $$\ce{NaOH}$$ is $$\frac{2.0 mmol}{80 mL} = 0.025 M$$. The concentration $$\ce{[H+] = 10^{-14}/0.025 = 4 10^{-13}M}$$. So $$p$$H = $$12.4$$. You see that the pH has fallen from $$13.0$$ to $$12.4$$ for more than half the trip to the equivalencce point. A change from $$13.0$$ to $$12.4$$ is rather small. The point of the curve is still nearly at its initial value. The gradient of pH between the beginning and here is small.

If you take the same calculation with an addition of $$45 mL~ \ce{ NaOH}~ 0.1 M$$, or $$4.5 mmol$$, the solution contains still $$5 mmol - 4.5 mmol = 0.5 mmol ~\ce{NaOH}$$ in $$95 mL$$. The concentration in $$\ce{NaOH}$$ is $$\frac{0.5 mmol}{95 mL} = 5.26·10^{-3} M$$. $$\ce{[H^+] = 1,9·10^{-12}}$$, and $$p$$H = $$11.7$$ We are near the equivalence point but the pH has not decreased a lot. It is still $$11.7$$. As you see the gradient (or the slope) of the curve is not important in the part of the curve between $$0 mL$$ and $$45 mL$$.

And suddenly between $$45 mL$$ and $$50 mL$$, the $$p$$H drops from $$11.7$$ to $$7$$. This is a huge change in just $$5 mL$$.

If you want you can calculate one or two other points along this titration. Try for example $$49 mL$$. You will be surprised ! It still gives a rather high $$p$$H value.

The titration curve is a deception of mathematics to our eyes. Look at the y-axis carefully. What do you have on the y-axis? A log function operating on hydrogen ion concentration. A unit change in pH means that hydrogen ions concentration has changed by a factor of 10. As more and more hydroxide ions are removed from the solution during the titration with HCl it takes lesser and lesser volume of acid to change the pH, hence a steep drop in your titration curve.

The concentration of hydrogen ions decreases proportionally with the addition of base . You add x-moles of hydroxide ions, x-moles of hydrogen ions are consumed.