A 25.0 mL sample of 0.100 M propanoic acid ( HC 3 H 5 O 2 , Ka = 1.30× 10 5 ) is

Question

A 25.0 mL sample of 0.100 M propanoic acid ( HC3H5O2, Ka = 1.30× 105 ) is titrated with 0.100 M KOH.
Calculate the pH after the addition of 20.0 mL of theKOH.

Explanation / Answer

HC3H5O2 + OH-   =>   C3H5O2-   +H2O moles propanoic acid = 0.025 L*(0.1 mol/L) = 0.0025 mol HC3H5O2 moles hydroxide = 0.02 L *0.1 mol/L) = 0.002 mol OH- react to from 0.002 mol C3H5O2-    and leave 0.0005mol HC3H5O2 V = 0.025 + 0.02 = 0.045 L [C3H5O2-] = 0.002 mol/ (0.045 L) = 0.0444444444 mol/L [HC3H5O2] = 0.0005 mol/0.045 L = 0.0111111111 mol/L           HC3H5O2 +   H2O   C3H5O2- +   H3O+ init       0.0111                                0.0444 change    -x                                          +x              +x equil    0.0111 -x                       0.0444 +x              x Ka = [C3H5O2-][H3O+]/[HC3H5O2] = (x)*(0.0444 + x)/(0.0111 - x) =1.3e-5 x2 + 0.0444x = 1.44444444e-7 - 1.3e-5 x x2 + 0.0444574444x - 1.44444444e-7 = 0 Solve via quadratic formula x = 3.25 e-6 mol/L x = [OH-] pH = 14 + log[OH-] = 14 + log(3.25e-6) = 8.51

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---