Tuesday
October 21, 2014

Homework Help: Chemistry

Posted by Zell on Thursday, October 4, 2012 at 3:09am.

Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq)

(a) What is the initial pH of the 0.20 M C6H5COOH(aq)?

(b) What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?


(c) What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?


(d) Calculate the pH at the halfway point.

(e) What volume of 0.30 M KOH(aq) is required to reach the stoichiometric point?


(f) Calculate the pH at the stoichiometric point.

---------attempted solutions:

C6H5COOH + H2O ==> C6H5COO^- + H3O^+
.2*.3
-x +x +x

[x^2]/[.6-x] = 6.5E-5
.00000039-6.5E-5x-x^2=0
x=.000593 moles
30 mL = .003L
[H+] = x/.003
pH=-log[H+]=0.704

Is this right? It seems really low...

b)
.20M * .03L = .006 mol C6H5COOH
.30 * .015L = .0045 mol KOH
.0045-.006 = .0015 excess

pKa C6H5COOH = 4.19

.03 + .015 = .045 L total volume

pH = pka + log([base]/[acid]) =
4.19 + log( (.0045/.045)/(.006/.045)) =
4.06

Is this right?

I haven't gotten to the other parts of the problem yet.

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