± Titration of Weak Acid with Strong Base A certain weak acid, HA, with a K a va

Question

± Titration of Weak Acid with Strong Base

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 7.00 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.

pH=??

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 64.0 mL ?

Express the pH numerically to two decimal places.

pH=??

Explanation / Answer

Ka = 5.61 x 10^-6

pKa = -logKa

pKa = 5.25

part A)

HA     + OH- --------------> A- + H2O

7           3                         0         0

4            0                         3          0

pH = pKa + log [A- / HA]

pH = 5.25 + log (3 /4)

pH = 5.12

part B)

at equivalence point only salt remains

salt concentratio = 7 / 64 = 0.109 M

pH = 7 + 1/2 [pKa + log C]

pH = 7 + 1/2 [5.25 + log 0.109]

pH = 9.14

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