What is the pH at the equivalence point for the titration of 0.16M solutions of

Question

What is the pH at the equivalence point for the titration of 0.16M solutions of the following acids and bases.
NaHSO3 and NaOH

This is what I did so far but it's incorrect and I'm not sure why.

0.16 mol Ba(OH)2 (weak base)
0.16 mol HBR (strong acid)

Ba(OH)2 = 0.16 mol/2 L = 0.08M

kB of Ba(OH)2 = 5.0x 10-3
ka = (1x10-14) / (5.0x 10-3) = 2.0x10-12

2.0x10-12 = x2/0.08
x2 = 1.6x10-13
x= 4.0x10-7
pH = -log(x) = 6.40

However, this is wrong. Please point out where I'm making the mistake so I can get this problem right.

Explanation / Answer

The reaction is,

Ba(OH)2 + HBr ------> Ba(OH)Br + H2O

Ba(OH)Br + HBr -------> BaBr2 + H2O

The solution initially contains 0.16 / 2 = 0.08 M each of Ba(OH)2 and HBr

Concentration of Ba(OH)Br produced from the stoichiometry of the 1st reaction = 0.08 M

Now we do not have any HBr left out for the second reaction.

Therefore, the solution now contains 0.08 M Na(OH)Br

Na(OH)Br is an amphoteric salt which can act as an Acid as well as a Base.

pOH of this Amphoteric salt (Na(OH)Br) is given by,

pOH = 0.5 x (pKb1 + pKb2)

pKb1 = -2.02 (Source: Wikipedia)

pKb2 = 2.3

pOH = 0.5 x (2.3 - 2.02) = 0.14

(Check the values of pKb1 and pKb2 with your textbook)

pH = 14 - pOH = 13.86

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