Question

Calculate the pH at the equivalence point in titrating 0.090 M solutions of each of the following with 0.064 M NaOH.




(a) hydroiodic acid (HI)

pH =  



(b) hydrazoic acid (HN₃), K_a = 1.9e-05

pH =  



(c) arsenous acid (H₃AsO₃), K_a = 5.1e-10

pH =  

Answer 1

(a) 1 mole of HI gets neutralized with 1 mole of NaOH

excess HI left = 0.090 - 0.064 = 0.026 M

HI is a strong acid dissociates completely into H+ and I-

pH = -log[H+] = -log(0.026) = 1.58

(b) Here we get salt of acid

remaining [HN3] = 0.09 - 0.064 = 0.026 M

formed [N3-] = 0.064 M

Using,

pH = pKa + log(base/acid)

pKa = -log[Ka] = -log(1.9 x 10^-5) = 4.72

So,

pH = 4.72 + log(0.064/0.026)

     = 5.11

(c) arsenous acid

concentration of [H2AsO3-] = 0.064 M

concentration of [H2AsO3] = 0.090 - 0.064 = 0.026 M

pH = pKa + log(base/acid)

     = 9.29 + log(0.064/0.026)

     = 9.68