Question

Calculate the mass of sodium hydroxide that must be added to 1.00 L of 1.00 M HC₂H₃O₂ to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

Short answer

40 g of NaOH must be added to double the pH of 1 L of 1 M acetic acid solution.

Step-by-step solution

Calculating pH of Acetic Acid

As acetic acid is a weak acid so it does not completely ionize in a solution. Hence, for calculating its pH, the acid dissociation constant has to be used.

$\begin{array}{l}{\text{H}}^{\text{+}}\text{=}\sqrt{{\text{K}}_{\text{a}}\text{×C}}\\ {\text{H}}^{\text{+}}\text{=}\sqrt{{\text{1.8×10}}^{\text{-5}}\text{×1}}\\ {\text{H}}^{\text{+}}{\text{=4.24×10}}^{\text{-3}}\end{array}$

From H⁺ ion concentration pH can be easily calculated:

$\begin{array}{l}{\text{pH=-logH}}^{\text{+}}\\ {\text{pH=-log(4.24×10}}^{\text{-3}}\text{)}\\ \text{pH=2.37}\end{array}$

Calculating Concentration of NaOH

On adding a strong base (NaOH), acetic acid will form its conjugate base resulting in a buffer solution. The pH of a buffer solution is calculated using the Henderson-Hasselbalch equation.

${\text{pH=pK}}_{\text{a}}\text{+log}\frac{\text{[Conjugate\hspace{0.17em}base]}}{\text{[Weak\hspace{0.17em}acid]}}$

As said in the question pH was double after adding NaOH hence,

$\begin{array}{c}{\text{2×2.37=-log(1.8×10}}^{\text{-5}}\text{)+log}\frac{\text{[Conjugate\hspace{0.17em}base]}}{\text{[Weak\hspace{0.17em}acid]}}\\ \text{4.74=4.74+log}\frac{\text{[Conjugate\hspace{0.17em}base]}}{\text{[Weak\hspace{0.17em}acid]}}\\ \text{log}\frac{\text{[Conjugate\hspace{0.17em}base]}}{\text{[Weak\hspace{0.17em}acid]}}\text{=0}\\ \frac{\text{[Conjugate\hspace{0.17em}base]}}{\text{[Weak\hspace{0.17em}acid]}}\text{=1}\\ \text{[Conjugate\hspace{0.17em}base]=[Weak\hspace{0.17em}acid]}\end{array}$

Concentration of conjugate base is equal to the concentration of NaOH. So, the concentration of NaOH will also be equal to weak acid to double the pH.

[NaOH] = 1mol/L

Calculating Mass of NaOH

For calculating mass,

$\begin{array}{l}\text{Mole=}\frac{\text{Mass}}{\text{MolarMass}}\text{=1}\\ \text{Mass=Mole×MolarMass}\\ \text{Mass=1×40=40g}\end{array}$

Hence, 40 g of NaOH must be added to double the pH of 1 M acetic acid solution.