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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,. .. &amp; you had 3 sigfigs visible everywhere but in the &quot;50 ml&quot; you may need to look to see if it had any more than just the 1 sigfig showing in the &quot;50 ml&quot;, &amp; 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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- Spring '07