To do these you must be able to recognize which anions/cations act as bases/acids in aqueous solution.
b part. NaClO is the salt. The ClO^- hydrolyzes in water as follows:
.........ClO^- + HOH --> HClO + OH^-
initial.0.0444.............0.....0
change...-x................x.....x
equil..0.044-x..............x....x
Kb for ClO^- = (Kw/Ka for HClO) = (HClO)(OH^-)/(ClO^-)
Substitute from ICE chart above and solve for x = (OH^-) and convert to pH.
The c part is just like the b part. For the a part, neither anion nor cation is hydrolyzed; therefore, the pH = 7.
Note: There is nothing in the problem that tells you how to handle the concn of the salt at the equivalence point but it is 0.0444 M in this instance. I don't know what level you are but it seems a little advanced for beginners.
You do it this way. Take ANY volume acid, calculate the volume of the 0.08M base it will take to neutralize it, then moles salt (M x L) divided by TOTAL volume. It makes no difference what volume you take it will always end up 0.0444M for the 0.1M/0.08M case.
HClO2 and NaOH gives you NaClO2
Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH.
a)hydrobromic acid (HBr)
b)chlorous acid (HClO2)
c)benzoic acid (C6H5COOH)
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