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A 30.0-mL sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. 1. Calculate the pH at 5 mL of added base. 2. Calculate the pH at one-half of the equivalence point. (Equivalence point is 8.96) 3. Calculate the pH at 20 mL of added base.

Victoria Fowler

in Chemistry

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Kathy Robinson on December 4, 2018

Thank you for posting your question here on brainly. I hope the answer will help you. Feel free to ask more questions here.moles of acid = 0.165 x 0.0300 L= 0.00495 to get the Ka we know thet pH at the equivalence point is 8.96 pOH = 14 - 8.96 = 5.04 [OH-]= 10^-5.04 = 9.12 x 10^-6 M = [CH3CH2COO-] Kb = Kw/Ka = (x 9.12 10^-6)^2 / 0.165 = 5.04 x 10^-10 Ka = Kw/Kb = 1.0 x 10^-14 / 5.04 x 10^-10=1.98 x 10^-5 pKa = 4.70 moles of OH - added = 0.005 L x 0.300 M= 0.0015 moles acid in excess = 0.00495 - 0.0015 = 0.00345 mole CH3CH2COO formed = 0.0015 pH = 4.70 + log 0.0015/ 0.00345=4.34 at the half-equivalence point pH = pKa = 4.70 moles of OH - added = 0.020 L x 0.300 M=0.00600 mol OH - in excess = 0.00600 - 0.00495=0.00105 total volume = 0.050 L [OH-]= 0.00105 / 0.050 L=0.0210 M pOH = 1.68 pH = 14 - 1.68=12.3


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