# chemsitry

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1)100ml sample of solution that is 0.2M in both Naf and Hf has 4.0 ml of 1.0M hcl added to it. calculate the ph change of this soultion .ka =7.2*10^-4

2)40 ml of 0.32M benzoic acid if titrated with 60 ml of 0.2 M naoh. clacilate the ph of the resulting solution at the equivalence point. ka =4.9*10^-9

• chemsitry -

Don't get lazy in chemistry. You see what difference there is between a 1m and a 1M solution. By the same token, Hf means heat formation and not HF which I assume you mean hydrofluoric acid. hcl and Naf mean nothing. Start your sentences with a capital letter. To your credit I can tell where the sentences end because you used a period.
100 mL x 0.2M NaF = 20 millimoles.
100 mL x 0.2M HF = 20 mmoles.
4.0mL x 1.0M HCl = 4 mmoles.

.........F^- + H^+ ==> HF
initial..20.....0......20
add............4.0.........
change..-4.0..-4.0....+4.0
equil...16.0...0......4.0

Use the Henderson-Hasselbalch equation, Substitute the equilibrium line from the ICE chart above and solve for pH.
Post your work if you need additional help.

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