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When 100.0 mL of weak acid HA were titrated with 0.09381 M RbOH, 27.36 mL were...

When 100.0 mL of weak acid HA were titrated with 0.09381 M RbOH, 27.36 mL were required to reach the equivalence point.

(a) Find the molarity of HA.

(b) What is the formal concentration of A- at the equivalence point?

(c) The pH at the equivalence point was 10.99. Find pKa for HA.

(d) What was the pH when only 19.47 mL of RbOH has been added?

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Homework Answers

Answer #1

a) at equivalence point HA moles = RbOH moles

Hence MV of acid = MV of base

M x 100 = 0.09381 x27.36

M of acid = 0.02567 M

b) A- moles = acid moles used = 0.02567 x 100/1000 = 0.002567

volume = 100 ml + 27.36 = 127.36 ml = 0.12736 L

[A-] = ( 0.002567/0.12736) = 0.02

c) pH = 10.99 , pOH = 14-10.99 = 3.01 , [OH-] = 10^ -3.01 = 0.00098 M

at equilibrium we have reaction   A- + H2O (l) <---> HA(aq) + OH- (aq)

[HA] = [OH-] =0.00098 M , [A-] = 0.02 ( nearly)

Kb = [HA][OH-]/[A-] = ( 0.00098) ^2 / ( 0.02) = 4.8 x 10^ -5

Ka = Kw/Kb = ( 10^ -14) / ( 4.8x10^-5) = 2.08 x 10^ -10

d) RbOH moles = 0.09381 x 19.47/1000 = 0.0018265 = A_ moles

HA moles left after reacting RbOH moles = 0.00257-0.0018265 = 0.00074

pH = pka + log [A-]/[HA]

pH = 9.55 + log [ 0.0018265 /0.00074)

   = 9.94

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