# How can I calculate the titration of a weak acid and a strong base?

Apr 20, 2015

There are some critical pH values you will need to calculate

• At the very beginning
• At the equivalence point
• Between neutralization and the equivalence point
• After the equivalence point

Assume that you are titrating 25 mL of 0.1 mol/L $\text{HA}$ (K_"a" = 10^-5) with 0.1 mol/L $\text{NaOH}$.

At the beginning

You know from your ICE table that

K_"a" = (["H"^+]["A"^-])/(["HA"]) = x^2/(0.1-x) = 10^-5

$x = 0.001$ and $\text{pH} = 3.00$

At the equivalence point

You know that the equivalence point will be at 25 mL of $\text{NaOH}$.

The $\text{HA}$ will be completely converted to ${\text{A}}^{-}$ and in twice the volume. Its concentration will be 0.05 mol/L

${\text{A"^(-) + "H"_2"O" ⇌ "HA" + "OH}}^{-}$;

K_"b" = K_"w"/K_"a" = (1.00 × 10^-14)/(10^-5) = 10^-9

K_"b" = (["HA"]["A"^-])/(["A"^-]) = x^2/(0.05-x) = 10^-9

x = 3.16 ×10^-4 and $\text{pOH} = 4.50$. So $\text{pH} = 9.50 .$

At points between 0 mL and the equivalence point

Use the Henderson-Hasselbalch Equation.

Calculate the moles of $\text{HA}$ and of ${\text{A}}^{-}$ remaining, and insert them into the equation.

For example, at 12.5 mL, you will have added 0.001 25 mol of base. You will have neutralized 0.001 25 mol of $\text{HA}$ and formed 0.001 25 mol of ${\text{A}}^{-}$.

"pH" = "p"K"a" + log((["A"^-])/(["HA"])) = -log(10^-5) + log((0.001 25)/(0.001 25)) = 5.00

After the equivalence point

You will just be adding excess moles of ${\text{OH}}^{-}$.

Calculate the excess moles, divide by the volume to get the molarity. Then calculate the pOH and the pH.

For example, after 40 mL of base, you will have added 0.004 mol of ${\text{OH}}^{-}$, but 0.0025 mol will have reacted with the acid.

You will have 0.0015 mol of ${\text{OH}}^{-}$ in 65 mL of solution.

["OH"^-] = "0.0015 mol"/"0.065 L" = "0.023 mol/L", so

$\text{pOH} = 1.64$ and $\text{pH = 12.36}$.

Your calculated values should match the graph below. 