Question

25.00 cm3 of 1.000 M propionic acid (structure CH3CH2COOH ) ( Ka = 1.34×10−5 ) is titrated with 0.910 M NaOH. What is the pH of the solution a) before any NaOH is added? b) at half-neutralization? c) at the equivalence point? d) when 0.10 mL less than the volume of NaOH to reach the equivalence point is reached? e) when 0.10 mL more than the volume of NaOH to reache the equivalence point is reached? f) Use your data to construct a plot of pH versus volume NaOH added (x-axis = vol NaOH, y-axis is pH)

​Please do not forget to do part (f); just give me a basic idea of what the graph should look like. Thank you.

Answer 1

a)

pKa = -log (Ka) = -log (1.34 x 10^-5)

pKa = 4.87

pH = 1/2 (pKa - log C)

     = 1/2 (4.87 - log 1)

pH = 2.44

b)

At half - equivalence point :

pH = pKa

pH = 4.87

C)

millimoles of weak acid = 25 x 1 = 25

At equivalence point :

millimoles of acid = millimoles of base

25 = 0.910 x v

V = 27.47 mL

here salt only remains.

salt concentration = 25 / (25 + 27.47) = 0.476 M

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (4.87 + log 0.476)

pH = 9.27

d)

millimoles of base = 0.910 x 27.37 = 24.9

Acid +   base    ------------> salt

25         24.9                          0

0.0933    0                          24.9

pH = pKa + log [salt / acid]

     = 4.87 + log [24.9 / 0.0933]

pH = 7.30

e)

millimoles of base = 25.089

here strong base remains.

base concentration = 0.0887 / 25 + 24.57 = 1.79 x 10^-3 M

pOH = -log (1.79 x 10^-3)

pOH = 2.75

pH = 11.25

the graph is like this :