Titration Of A Weak Acid With A Strong Base - Chemistry LibreTexts

Example \(\PageIndex{4}\): After adding 25 mL of 0.3 M NaOH

Find the pH after the addition of 25 mL of NaOH.

SOLUTION

• The millimoles of OH- added in 25 mL:\(25mL* \dfrac{.3mmol OH^{-}}{1mL} =7.5 mmol OH^{-}\)

HF	H2O	H3O+	F-
Initial	7.5 mmol	0 mmol	-	0 mmol
Add	0 mmol	+7.5 mmol	-	0 mmol
Change	-7.5 mmol	-7.5 mmol	-	7.5 mmol
Equilibrium	0 mmol	0 mmol	-	7.5 mmol

This is the equivalence point of the titration. We know this because the acid and base are both neutralized and neither is in excess. To find the concentrations we must divide by the total volume. This is the initial volume of HF, 25 mL, and the addition of NaOH, 25 mL. Therefore the total volume is 25 mL + 25 mL = 50 mL

Concentration of F-:\(\dfrac{7.5 mmol F^{-}}{50 mL}=0.15M\)

However, to get the pH at this point we must realize that F- will hydrolyze. An ICE table for this reaction must be constructed

HF	H2O	H3O+	F-
Initial	0.15 M	-	0 M	0 M
Change	- X	-	+ X	+X
Equilibrium	0.15 - X	-	X M	X M

In this reaction the F- acts as a base. Therefore we must obtain the kb value instead of the ka value.

\(k_{b}= \dfrac{k_{w}}{k_{a}}\)

\(k_{b} = \dfrac{1.0\times 10^{-14}}{6.6\times 10^{-4}}\)

\(k_{b}=1.515\times 10^{-11}\)

Now that we have the kb value, we can write the ICE table in equation the equation form

\(1.515\times 10^{-11} \dfrac{x^{2}}{.15-x}\)

Manipulating the equation to get everything on one side yields

\(0= x^{2} + 1.515 \times 10^{-11}x -2.2727\times 10^{-12}\)

Now this information is plugged into the quadratic formula to give

\(x = \dfrac{-1.515\times 10^{-11} \pm \sqrt{(-1.515\times 10^{-11})^2 - 4(1)(-2.2727\times 10^{-12})}}{2}\)

The quadratic formula yields x=1.5075\times 10-6 and -1.5075\times 10-6 . However the negative value can be ruled out because concentrations cannot be zero.

Therefore to get the pOH we plug the concentration of OH- into the equation pH=-log(1.5075\times 10-6) and get pOH=5.82. To get the pH we minus the pOH from 14.

pH=14 - 5.82

pH= 8.18