Question

Calculate the pH at the equivalence point in titrating 0.089 M solutions of each of the following with 0.061 M NaOH.




(a) hydrochloric acid (HCl)

pH =  



(b) propionic acid (HC₃H₅O₂), K_a = 1.3e-05

pH =  



(c) phenol (HC₆H₅O), K_a = 1.3e-10

pH =  

Answer 1

a) NaOH + HCl ------> NaCl + H2O

strong base and strong acid titration so at equivalence point.

pH = 7.0

b) titration of weak acid and strong base

C3H5COOH + NaOH --------------> C3H5COONa + H2O

[C3H5COONa] = 0.061 M

[C3H5COOH] = 0.089 - 0.061 = 0.028 M

mixture of these two act as acidic buffer

pH = pKa + log [C3H5COONa] / [C3H5COOH]

pKa = - log Ka = - log [1.3 x 10^-5] = 4.89

pH = 4.89 + log [0.061] / [0.028]

pH = 5.23

c) titration of weak acid and strong base

C6H5OH + NaOH ----> C6H5ONa + H2O

pKa = - log Ka = - log [1.3 x 10^-10] =9.89

[C6H5ONa] = 0.061

[C6H5OH] = 0.089 - 0.061 = 0.028 M

pH = pKa + log [C6H5ONa] / [C6H5OH]

pH = 9.89 + log [0.061] / [0.028]

pH = 10.23