****please help..i was able to get part A but i am getting the wrong answers for

Question

****please help..i was able to get part A but i am getting the wrong answers for part b.c.and d.

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 3.17-M HCl to 430. mL of each of the following solutions.

Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.

b) 0.183 M C2H3O21-

c) 0.183 M HC2H3O2

d) a buffer solution that is 0.183 M in each C2H3O21- and HC2H3O2

Explanation / Answer

b) 0.183 M C2H3O2^-

pH before mixing = -log[H+] = -log(3.17) = 0.50

moles of HCl = molarity (M) x volume (L) = 3.17 x 0.01 = 0.0317 mols

moles of C2H3O2^- = 0.183 x 0.43 = 0.079 mols

we have 0.0317 mols of HC2H3O2

remaining C2H3O2- = 0.079 - 0.0317 = 0.047 mols

Molarity of HC2H3O2 = 0.0317/0.440 = 0.072 M

molarity of C2H3O2^- = 0.047/0.44 = 0.107 M

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

      = 4.76 + log(0.107/0.072)

      = 4.93

So pH after mxing = 4.93

Change in pH = 4.93 - 0.5 = 4.43

c) 0.183 M HC2H3O2

Initial pH = 0.5

HC2H3O2 <==> H+ + AcO-

molarity of HC2H3O2 in total solution = 0.183 x 0.43/0.44 = 0.179 M

1.74 x 10^-5 = x^2/0.179

x = [H+] = 1.765 x 10^-3 M

Total [H+] = 1.765 x 10^-3 + (3.17 x 0.01/0.44) = 0.074

pH = -log(0.074) = 1.13

Change in pH = 1.13 - 0.5 = 0.63

c) pH before mixing = 0.5

moles of HCl = 3.17 x 0.01 = 0.0317 mols

final molarity of C2H3O2^- = [0.183 x 0.43 - 0.0317]/0.44 = 0.107 M

final molarity of HC2H2O2 = [0.183 x 0.43 + 0.0317]/0.44 = 0.251 M

Hendersen-Hasselbalck equation,

pH = 4.76 + log(0.107/0.251)

     = 4.39

Change in pH = 4.39 - 0.5 = 3.89

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