The Ka for vitamin C (also known as ascorbic acid, C6H8O6) is 8.0 x 10^-5. If 15 g of vitamin C is dissolved in 1.00 L of water, what is the pH of the solution at equilibrium
The Correct Answer and Explanation is:
To find the pH of a vitamin C solution at equilibrium, we first need to understand the dissociation of ascorbic acid (C6H8O6) in water. Ascorbic acid is a weak acid and dissociates according to the following equilibrium reaction:C6H8O6 (aq)⇌H+ (aq)+C6H7O6- (aq)\text{C6H8O6 (aq)} \rightleftharpoons \text{H+ (aq)} + \text{C6H7O6- (aq)}C6H8O6 (aq)⇌H+ (aq)+C6H7O6- (aq)
The equilibrium constant expression for this dissociation is given by:Ka=[H+][C6H7O6-][C6H8O6]K_a = \frac{[\text{H+}][\text{C6H7O6-}]}{[\text{C6H8O6}]}Ka=[C6H8O6][H+][C6H7O6-]
Step 1: Convert the given mass of vitamin C to moles
We are given 15 g of vitamin C dissolved in 1.00 L of water. To find the number of moles of vitamin C, we use its molar mass. The molar mass of ascorbic acid (C6H8O6) is:M=6(12.01)+8(1.008)+6(16.00)=176.12 g/molM = 6(12.01) + 8(1.008) + 6(16.00) = 176.12 \, \text{g/mol}M=6(12.01)+8(1.008)+6(16.00)=176.12g/mol
Now, calculate the number of moles of vitamin C:Moles of C6H8O6=15 g176.12 g/mol=0.0852 mol\text{Moles of C6H8O6} = \frac{15 \, \text{g}}{176.12 \, \text{g/mol}} = 0.0852 \, \text{mol}Moles of C6H8O6=176.12g/mol15g=0.0852mol
Step 2: Write the ICE table for the dissociation
Let’s assume that the acid dissociates to a small extent. Let xxx be the concentration of H+ ions that dissociate from the acid.C6H8O6H+C6H7O6-Initial concentration0.0852 M00Change−x+x+xEquilibrium concentration0.0852−xxx\begin{array}{|c|c|c|c|} \hline & \text{C6H8O6} & \text{H+} & \text{C6H7O6-} \\ \hline \text{Initial concentration} & 0.0852 \, \text{M} & 0 & 0 \\ \text{Change} & -x & +x & +x \\ \text{Equilibrium concentration} & 0.0852 – x & x & x \\ \hline \end{array}Initial concentrationChangeEquilibrium concentrationC6H8O60.0852M−x0.0852−xH+0+xxC6H7O6-0+xx
Step 3: Set up the equilibrium expression
Using the Ka value of 8.0 x 10^-5, we can substitute the equilibrium concentrations into the Ka expression:Ka=x⋅x0.0852−x=8.0×10−5K_a = \frac{x \cdot x}{0.0852 – x} = 8.0 \times 10^{-5}Ka=0.0852−xx⋅x=8.0×10−5
This simplifies to:x20.0852−x=8.0×10−5\frac{x^2}{0.0852 – x} = 8.0 \times 10^{-5}0.0852−xx2=8.0×10−5
For weak acids, the value of xxx (H+ concentration) will be small compared to the initial concentration of the acid. So, we can approximate 0.0852−x≈0.08520.0852 – x \approx 0.08520.0852−x≈0.0852, simplifying the equation to:x20.0852=8.0×10−5\frac{x^2}{0.0852} = 8.0 \times 10^{-5}0.0852×2=8.0×10−5
Now, solve for xxx:x2=(8.0×10−5)×0.0852=6.816×10−6x^2 = (8.0 \times 10^{-5}) \times 0.0852 = 6.816 \times 10^{-6}x2=(8.0×10−5)×0.0852=6.816×10−6x=6.816×10−6=2.61×10−3 Mx = \sqrt{6.816 \times 10^{-6}} = 2.61 \times 10^{-3} \, \text{M}x=6.816×10−6=2.61×10−3M
Step 4: Calculate the pH
The concentration of H+ ions is 2.61×10−3 M2.61 \times 10^{-3} \, \text{M}2.61×10−3M. The pH is given by:pH=−log[H+]\text{pH} = -\log[\text{H+}]pH=−log[H+]
Substituting the value of xxx:pH=−log(2.61×10−3)=2.58\text{pH} = -\log(2.61 \times 10^{-3}) = 2.58pH=−log(2.61×10−3)=2.58
Final Answer
The pH of the vitamin C solution at equilibrium is 2.58.
