Microsoft - pOH = 5.284 pH = 14 - pOH = 8.716 your answer...

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Calculate the pH at the equivalence point for a titration of acetic acid and NaOH? If Ka is 1.85x10-5 for acetic acid, calculate the pH at one half the equivalence point and at the equivalence point for a titration of 50mL of 0.100 M acetic acid with 0.100 M NaOH. ANSWER: pH @ one half the equivalence point equals th pKa if the Ka = 1.85e-5,. .. the pKa = 4.733 your answer(3sigfigs): pH @ 1/2 reaction is = 4.733 At the end, a titration of 50mL of 0.100 M acetic acid with 0.100 M NaOH, you would think it might produce 0.1 Molar sodium acetate. .. but mixing 50 ml base with 50 ml acid dilutes the salt in half to 0.0500 Molar sodium acetate C2H3O2)-1 in water => HC2H3O2 & OH- 0.0500 -x . ........ Kb = Kwater / Ka = 1e-14 / 1.85e-5 = 5.405 e-10 Kb = [HC2H3O2] [OH-] / [C2H3O2)-1] 5.405 e-10 = [HC2H3O2] [OH-] / [C2H3O2)-1] 5.405 e-10 = [x] [x] / [0.0500] x = [OH-} = 5.198 e-6
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Unformatted text preview: pOH = 5.284 pH = 14 - pOH = 8.716 your answer (3sigfigs): 8.716 the problem though is that I gave you 3 sig fig answers,. .. & you had 3 sigfigs visible everywhere but in the "50 ml" you may need to look to see if it had any more than just the 1 sigfig showing in the "50 ml", & round off your answers accordingly PH at the equivalence point in the titration of acetic acid with sodium hydroxide when given Kb of conj. base? What is the pH at the equivalence point in the titration of 10.0mL of 0.32 M CH3CO2H with 15.0mL of 0.213 M NaOH? For CH3COO-, Kb=5.6*10^-10 ANSWER: at the equivalence point moles of NaOH added equals moles of acetic acid present n NaOH = n CH3CO2H = M*V = 0.213*0.015= 3.20*10^-3 mol = n CH3COO- formed final solution volume = 25.0 mL = 2.5*10^-2 L [CH3COO-] = n/V = 3.20*10^-3 / 2.5*10^-2 = 0.128 M acetate is basic [OH-] = square root(5.6*10^-10*0.128) = 8.47*10^-6 M pH = 8.93...
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Microsoft - pOH = 5.284 pH = 14 - pOH = 8.716 your answer...

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