5. a) What is the pH of a 0.35 M solution of benzoic acid, HC 7 H 5 O 2 ? The K

Question

5. a) What is the pH of a 0.35 M solution of benzoic acid, HC7H5O2? The Ka for benzoic acid is 6.5 x 10-5. Show the reaction and all calculations.

b) What is the percent ionization of benzoic acid in the solution in part (a)?

c) When solid sodium benzoate is added to the solution in part (a) the percent ionization of benzoic acid ______________ . (increases / decreases / stays the same)

d) When solid sodium benzoate is added to the solution in part (a) the pH of the solution ______________ . (increases / decreases / stays the same)

e) What is the pH of a solution that is a mixture of 0.35 M benzoic acid and 0.15 M sodium benzoate?

f) What is the new pH if 0.05 moles of sodium hydroxide are added to 500 mL of the solution from part (e) on the previous page?

g) What is the pH when 0.05 moles of sodium hydroxide are added to 500 mL of pure water? Briefly explain the difference in your answers to parts (f) and (g).

Plese show detailed work on e, f, and g.

Explanation / Answer

a)

Ka = (H30)+ (C7H5O2)- / (HC7H5O2)
PH= - LOG(H30)+

(H)+= 10-pH

6.5*10-5 = (10(-pH))2/ 0.35

10-pH=4.77*10-3

-pH=log(4.77)-3=0.678-3=-2.32

pH=2.32

b)

Ka=x^2/0.350= 6.5 *10^5
x^2= 2.275 * 10^-5
x=4.76* 10^-3 = 0.00476 M this is how much the mole per liter of benzoic acid that ionize

the % ionization = 0.00476M/0.350M * 100 %= 1.362 %

c) decrease.

The effectiveness of sodium benzoate as a preservative increases with decreasing pH because the ratio of undissociated (i.e., free) benzoic acid to ionized benzoic acid increases as the pH decreases.

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