Asked by DANNY
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)
a)hydrobromic acid (HBr)
b)chlorous acid (HClO2)
c)benzoic acid (C6H5COOH)
Answers
Answered by
DrBob222
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
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
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