Homework Help: Chemistry

Posted by Dubs on Thursday, April 22, 2010 at 1:07am.

A few questions I don't really get and need to see the work for

A 50.0 mL sample of 0.55 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.51 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 multiplied by 10-5.

What is the pH of a solution resulting from the addition of 18.0 mL of 0.15 M HCl to 50.0 mL of 0.12 M NH3? Assume the volumes are additive. Kb of NH3 = 1.8 multiplied by 10-5.

What ratio [CH3NH3+]/[CH3NH2] is needed to prepare a buffer solution with a pH of 8.20 from methylamine, CH3NH2, and methylammonium chloride, CH3NH3Cl? Kb of CH3NH2 = 3.7 multiplied by 10-4.

What is the pH of a solution resulting from the addition of 18.0 mL of 0.15 M HCl to 50.0 mL of 0.12 M NH3? Assume the volumes are additive. Kb of NH3 = 1.8 multiplied by 10-5.

What is the molarity of a methylamine solution, CH3NH2(aq), that has the same pH as 0.0773 M NH3(aq)?

Chemistry - DrBob222, Thursday, April 22, 2010 at 11:35am
A few comments.
1. Five problems in one post-- four too many.
2. No work at all does not engender help and "I don't get it and need to see the work" is not a high incentive. Instead, post one problem and tell us what you don't understand. Surely you understand some things. Telling us what you know and what you don't understand helps us know where to start with the problem. We can help you better if we know how to explain it. Thanks.
3. Some of the rationale for the above comes from the fact that it can take up to 30 minutes each to type in a problem with all of the exponents etc and that can easily be wasted time if you already know that part of our explanation. Again, thanks.
Chemistry - DrBob222, Thursday, April 22, 2010 at 11:40am
The secret to doing these titration problems is in knowing what you have at certain points in the titration. For benzoic acid titrated to the equivalence point, the acid has been converted to the salt and one has sodium benzoate. Salts, as you know, hydrolyze. If we call benzoic acid HBz and the benzoate anion Bz, then
Bz^- + HOH ==> HB + OH^-
Kb for Bz^- = (Kw/Ka) = (HBz)(OH^-)/(Bz^-).
Substitute Kw, Ka, (Bz = moles/L) and (HBz)=(OH^-)=x. Solve for x, convert to pOH, then pH.

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