Question

A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH.

A solution is made by titrating 44.17 mmol (millimoles) of HA and 0.09 mmol of the strong base.

Then, 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 41.7 mL ?

Answer 1

We have to use ICE table as well as Henderson-Hasselbalch equation of buffers both here.

R......................HA....+........OH........<==> A-............+ ....H2O

I.......................44.17.........0.09..................0.........................

C.....................-0.09.........-0.09...............+0.09.....................

E......................44.08........0....................0.09

pH = pKa + log[salt/acid] = -logKa + log[salt/acid] = 5.25 + log[0.09/44.08] = 2.56

pH = 2.56

now at equivalence point all the acid is converted into salt and again salt undergoes hydrolysis

A^- concentration = 44.17 mmol / 41.7 mL = 1.059 M

the pH is determined by the hydrolysis of the A- the hydrolysis constant is Kw/Ka = 1.78 X 10^-9

so A- +H₂O ------ HA + OH^-

1.059 M -x amount ionized will give x amount of A and x amount of OH^-

x²/ 1.059 = 1.78 X 10^-9

X² = 1.8877 x 10^-9

X = OH^- = SQRT ( 1.8877 x 10^-9 ) = 0.00004344767

pOH = -log[OH^-] = -log[0.00004344767] = 4.362

pH = 14- pOH = 14 - 4.362 = 9.64

Final answer: pH = 9.64

Hope this helped you!

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