Question

A buffer solution contains 0.33 mol of hypoiodous acid (HIO) and 0.75 mol of sodium hypoiodite (NaIO) in 4.50 L.
The K_a of hypoiodous acid (HIO) is K_a = 2.3e-11.



(a) What is the pH of this buffer?

pH =  


(b) What is the pH of the buffer after the addition of 0.26 mol of NaOH? (assume no volume change)

pH =  


(c) What is the pH of the original buffer after the addition of 0.08 mol of HI? (assume no volume change)

pH =  

Answer 1

[HIO] = 0.33 mol

[NaIO] = 0.75 mol

Ka of HIO = 2.3x 10^-11

a)   pH = - log Ka + log [NaIO]/[HIO]

           = - log (2.3x 10^-11) + log ( 0.75/0.33)

           = 11

pH = 11

b) pH of the buffer after the addition of 0.26 mol of NaOH

     HIO        + NaOH         ------------>   NaIO + H2O

   0.33 mol      0.26 mol                  0

------------------------------------------------------------------------------

0.33-0.26            0                         0.26 mol

= 0.07 mol   

Hence,

[HIO] = 0.07 mol

[NaIO] = 0.75 mol + 0.26 mol = 1.01 mol

pH    = - log Ka + log [NaIO]/[HIO]

           = - log (2.3x 10^-11) + log ( 1.01/0.07)

           = 11.8

pH = 11.8

c) pH of the original buffer after the addition of 0.08 mol of HI

NaIO     +    HI         ------------>   HIO      + NaI

   0.75 mol        0.08 mol                  0

-------------------------------------------------------------------------------------

0.75-0.08            0                            0.08 mol

= 0.67 mol   

Hence,

[HIO] = 0.08 mol + 0.33 mol = 0.41 mol

[NaIO] = 0.67 mol

pH    = - log Ka + log [NaIO]/[HIO]

           = - log (2.3x 10^-11) + log ( 0.67/0.41)

           = 10.85

pH = 10.85