Answer to Question #83534 in General Chemistry for meghan

Question

A buffer is created by dissolving 0.450 mol of sodium ascorbate and 0.500 mol ascorbic acid in enough water to create 1.00L of solution. The Ka of ascorbic acid is 7.9 x 10-5. what is the pH when 0.015 mol of HCl is added to the buffer?

Answer 1

Answer on Question #83534, Chemistry/ General Chemistry

A buffer is created by dissolving 0.450 mol of sodium ascorbate and 0.500 mol ascorbic acid in enough water to create 1.00L of solution. The Ka of ascorbic acid is 7.9 x 10-5. what is the pH when 0.015 mol of HCl is added to the buffer?

Solution

To answer this question we should use Henderson-Hasselbalch equation:

pH=pKa+log([A^- ]/[HA] )

Where pH is the concentration of [H+]

pKa is the acid dissociation constant

[A^-]/[HA] is the ratio of the concentrations of the conjugate base and starting acid.

After addition:

Concentration of HCl added is 0.015 mol/1 L = 0.015 M

Initial concentration of sodium ascorbate is c(C₆H₇O₆Na) = 0.450 mol /1.00 L = 0.450 M

Initial concentration of ascorbic acid is c (C₆H₈O₆) = 0.500 mol/ 1.00 L =0.500 M

As we add a strong acid we make an assumption that all moles of conjugate base react with this acid to give a new concentration of [C₆H₇O₆^-] = 0.450-0.015 = 0.435 M and additional concentration of C₆H₈O₆ is formed [C₆H₈O₆] = 0.500+0.015=0.515 M

pH=pKa+log([A^- ]/[HA] )=-log⁡(7.9*10^-5)+log((0.450-0.015)/(0.500+0.015))=4.029

Answer: pH = 4.029