A certain weak acid, HA, with a Ka value of 5.61×106, is titrated with NaOH. Par

Question

A certain weak acid, HA, with a Ka value of 5.61×106, is titrated with NaOH.

Part: A A solution is made by titrating 8.00 mmol (millimoles) of HA and 3.00 mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places. I know Part A answer is 5.03. But I can't figure out Part B!

More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 34.0 mL ?

Express the pH numerically to two decimal places.

Please help! Thanks a bunch!!

Explanation / Answer

pKa = -log(Ka) = -log(5.61*10^-6) =5.2510

HA <->H+ + A-

Ka = [H+][A-]/[HA]

pH = pKa + log(A-/HA)

mmol of base = 3

after reaction...

mmol of acid left = 8-3 = 5mmol

mmol of conjguate formed = 0 + 3 = 3

so

pH = 5.2510 + log(3/5) = 5.029151

pH = 5.03

b)

equivalence point is reached...

Vtotal = 34 mL

mmol of acid = 8

[A-] = mmol / Vtotal = 8/34 = 0.235294 M

so

there is only A- in solution

A- + H2O <->HA + OH-

Kb = [HA][OH-]/[A-]

Kb = Kw/Ka = (10^-14)/(5.61*10^-6) = 1.78253*10^-9

in equilbirium

[HA] = [OH-]= x

[A-] = M-x = 0.235294 - x

substitute

Kb = [HA][OH-]/[A-]

1.78253*10^-9 = x*x/( 0.235294 - x)

x = 2.046*10^-5

[OH-.] =

pOH = -log(2.046*10^-5 ) = 4.68909

pH = 14-4.68909 = 9.31091 = 9.31

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