[CHM 112] additional problems

[Andrew J. Pounds]

On 04/09/2013 10:54 PM,  wrote:
> Will you explain to me how you came to set up the equation for 
> question 2 in chapter 17? "Calculate the concentration of C6H5COO- in 
> a solution that is 0.015M C6H5CooH and 0.015M HCl."

For this problem I knew I had the conjugate base of a weak acid (C_6 H_5 
COO^- ) and a strong acid (HCl) in solution.  This necessitated an 
equation that would have these two components (where the HCl, being a 
strong acid, is replaced by the hydronium ion) plus the weak acid (C_6 
H_5 COOH). The weak acid reaction equation, for which we also have a K_a 
, works nicely.

For the equilibrium expression it is simply

Ka = [products]/[reactants] = [C_6 H_5 COOH][H_3 O^+ ]/[C_6 H_5 COOH] = 
(x)(0.051+x)/(0.015-x)




-- 
Andrew J. Pounds, Ph.D.  (pounds at theochem.mercer.edu)
Associate Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627

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