Question

You are studying aspirin and its acid base properties. You find that 1.00L of a 0.500 M solution of aspirin has a pH of 1.86. You are interested in learning about the % dissociation in a buffered solution of aspirin so you make a new 1.0L solution containing 0.500moles of aspirin and 0.35 moles of the sodium salt of aspirin. What will the % disicociation be in the new buffered solution?

Answer 1

Answer on the question #58044, Chemistry / General Chemistry

Question:

You are studying aspirin and its acid base properties. You find that 1.00L of a 0.500 M solution of aspirin has a pH of 1.86. You are interested in learning about the % dissociation in a buffered solution of aspirin so you make a new 1.0L solution containing 0.500 moles of aspirin and 0.35 moles of the sodium salt of aspirin. What will the % disicociation be in the new buffered solution?

Solution:

From the data about pH and total concentration of the acid, one can calculate the equilibrium constant of aspirin dissociation.

$$K = \frac {[ H ^ {+} ] [ a s p ^ {-} ]}{[ H a s p ]} = \frac {[ H ^ {+} ] ^ {2}}{c ^ {0} (H a s p) - [ H ^ {+} ]} = \frac {1 0 ^ {- 2 p H}}{0 . 5 - 1 0 ^ {- p H}} = 3. 9 2 \cdot 1 0 ^ {- 4} m o l L ^ {- 1}$$

Then, we can use this constant to calculate the equilibrium concentration of hydrogen ions in buffer solution. For this we apply the following scheme:

$c^0 / \frac{mol}{L}$

0.5

0.35

$\Delta c / \frac{mol}{L}$

- $x$

+ x

+

$-x$

$-x$

$-L / \frac{mol}{L}$

0.5 - x

0.35 + x

x

$K = \frac{[H^{+}][asp^{-}]}{[Hasp]}$

$K = \frac{(0.35 + x)x}{(0.5 - x)}$

$x^{2} + (0.35 + K)x - 0.5K = 0$

Solving quadratic equation, we get $x$:

$$x = 5.58 \cdot 10^{-4}$$

Then, equilibrium concentration of hydrogen ions is:

$$[H^{+}] = 5.58 \cdot 10^{-4} \text{mol } L^{-1}$$

Dissociation percentage is the ratio between the concentration of hydrogen ions and concentration of undissociated species:

$$\alpha = \frac{[H^{+}]}{[\text{HA}]} \cdot 100\% = \frac{5.58 \cdot 10^{-4}}{0.5 - 5.58 \cdot 10^{-4}} \cdot 100\% = 0.11\%$$

Answer: $0.11\%$