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CHM135H1 (333)
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Department
Chemistry
Course
CHM135H1
Professor
Kris Quinlan
Semester
Summer

Description
Slide No. Notes 13 - half equivalence point half of acid added - equivalent concentration of HA and A- - at equivalent point, pH not 7 but greater than 7, because OH- present - EX. 40.00mL of 0.1000M HPr is titrated with 0.100 M NaOH. What is the pH after 24.00mL NaOH is added? - Ka(HPr) = 1.3 x 10^-5 - Same concentration so look at volume more acid, therefore acidic - Calculate [HPr] & [NaOH] after dilutionmixing up each other - [HPr] = 0.1000ML times 40.00mL divided by 65.00 mL = 0.06154 - [NaOH] = (0.100ML)(25.00mL) 65.00mL = 0.03846 - HPr + OH- -> water + Pr- - 0.06154M 0.03846M 0M - 0.02308M 0M 0.03846M - HPr + water -> Pr- + H3O+ - I: 0.02308M 0.03846M 0M - C: -x +x +x - E: 0.02308M-x 0.03846M+x x - Ka = [products][reactants] = (0.03846+x)(x)(0.02308- x) = 1.3 x 10^-5 - Assume x << 0.02308 M - X = 7.8 x 10 ^ -6 M - pH = 5.11
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