2, why is the pH and % dissociation of the HCl solutions different than that of

Question

2, why is the pH and % dissociation of the HCl solutions different than that of the acetic acid solutions of similar concentration? Explain the physical basis of these observations-don't just say "HCl is a stronger acid than HC H,0 (Hint: how does acetic acid dissociate in aqueous solution?) HC is a stronger acid than HCH,O" does a 3. In calculations with weak acids, we usually assume that the initial concentration is essentially unchanged by dissociation (i.e., HAinitia [HAl.) a. Based on your data, does this seem to be a valid assumption for all initial concentrations of acetic acid? Cite specific numerical evidence from your data table for acetic acid to justify your claims b. Does the corresponding assumption for base Bl) appear to he reasonable for all initial concentrations of sodium acetate? Cite specific numerical evidence from your data table for sodium acetate to justify your claims Calculate the percent error of your average experimental pK, and pk, values for acetic acid and acetate ion using the following accepted values: acetic acid, pK,-4.74; acetate ion, pK, 9.25. Show your work 4.

Explanation / Answer

Q2) HCl has acid dissociation constant ,Ka>1 So,it fully dissociates into its corresponding ions in solution.For acetic acid acid dissociation constant ,Ka=4.4*10^-6 <<<1 ,so it dissociates partially in solution and there exists an equilibrium between CH3COOH (aq),and CH3COO- (aq) , H+( aq) ions.

HCl(aq) ---->H+(aq) + Cl- (aq)

CH3COOH (aq) <-------> CH3COO- (aq) +H+( aq)

Q3)(a)

[HA]eqm=[HA]initial-x, where x=dissociated acetic acid

For negligible x, is the case when the dissociation is very very less for very small Ka.

This does not seem to be a valid claim from the values listed in the table as [CH3COOH]eq <<< [CH3COOH]intial

(b)

[A-]eqm=[A-]initial-x, where x=dissociated base acid

For negligible x, is the case when the dissociation is very very less for very small Kb (base dissociation constant).

[CH3COO-]eqm=[CH3COO-]initial-x

Q4)

% error =(difference in pka vaues/accepted vaue)*100=(expt value- accepted value)*100/accepted value

Ka=4.4*10^-6

pka=-log Ka=-log (4.4*10^-6)=5.4

% error=((5.4-4.74)/4.74)*100=13.9%

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