Question

when a 500 mg tables of vitamin c is dissolved in 50.0 mL water at 25 degrees Celcius, the result is a solution whose pH is 2.67 calculate the percent ionization

Answer 1

Step 1 - Discovering the total concentration of vitamin c

The molar concentration can be obtained by dividing the number of moles by the volume:

$\lbrack\text{ }\rbrack=\frac{m}{M\times V}$

In this equation, m represents the mass, M the molar mass and V the volume. For vitamin c, we have:

$\begin{gathered} m=500\text{ mg = 0.5 g} \\ \\ M=176\text{ g/mol} \\ \\ V=50\text{ ml = 0.05 L} \end{gathered}$

Substituting these values on the equation above:

$\lbrack\text{vitamin c}\rbrack=\frac{0.5}{176\times0.05}=\frac{0.5}{8.8}=0.0568\text{ mol/L}$

Step 2 - Finding the concentration of H+

Vitamin c is a monoprotic acid (I'll represent it by VitcH), and it will react with water to form H3O+:

$\text{VitcH + H}_2O\rightarrow H_3O^++Vitc^-$

The concentration of H3O+ that came from the ionization of vitamin c can be calculated by the pH:

$pH=-\log \lbrack H_3O^+\rbrack$

Since pH=2.67, we have:

$\lbrack H_3O^+\rbrack=^{}10^{-2.67}=0.0021\text{ mol/L}$

Step 3 - Finding the percent ionization

Now that we know how many vitamin c was dissolved and how many of it was ionized to form H3O+, we can set the following proportion:

$\begin{gathered} 0.0568\text{ mol/L ---- 100\%} \\ 0.0021\text{ mol/L---- x} \\ \\ x=\frac{100\times0.0021}{0.0568}=3.7\text{ \%} \end{gathered}$

The percent ionization is thus 3.7 %. Note that Vitamin c is a very weak acid.