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Submitted by geanowells on Wed, 12/09/2009 - 15:55

Consider the following acids and their dissociation constants.

HPO42-(aq) + H2O(l)  H3O+(aq) + PO43-(aq)  Ka = 2.2  10-13 
HCOOH(aq) + H2O(l)  H3O+(aq) + HCOO-(aq)  Ka = 1.8  10-4 

(a) Which is the weaker acid, HPO42- or HCOOH?

HPO42-
HCOOH   

(b) What is the conjugate base of HPO42-? (Type your answer using the format H3PO4 for H3PO4, (NH4)2CO3 for (NH4)2CO3, [NH4]+ for NH4+, and [Ni(CN)4]2- for Ni(CN)42-.)

(c) Which acid has the weaker conjugate base?

HPO42-
HCOOH   

(d) Which acid has the stronger conjugate base?

HPO42-
HCOOH   

For an acid, the higher the Ka value, the stronger the acid

This is not your example
H3PO4 + H2O ---> H3O+  +  H2PO4-
acid        bases    conj.acid  conj.base

HA  + H2O ---> H3O+  +  A- 
If Ka = 1 x 10-5, 'Ka' for the reverse reaction will be 1/(1 x 10-5) = 1 x 105. The larger this value, the stronger the conjugate base as it is more likely to remove H3O+ ions and move the equilibrium to the LHS

Submitted by kingchemist on Thu, 12/10/2009 - 04:36 Permalink

So what you're sayig is the higher the Ka value the stronger the acid, and the stronger the ka value the stronger the conjugte base as well?

For an acid, the higher the Ka value, the stronger the acid

This is not your example
H3PO4 + H2O ---> H3O+  +  H2PO4-
acid        bases     conj.acid   conj.base

HA   + H2O ---> H3O+  +  A-   
If Ka = 1 x 10-5, 'Ka' for the reverse reaction will be 1/(1 x 10-5) = 1 x 105. The larger this value, the stronger the conjugate base as it is more likely to remove H3O+ ions and move the equilibrium to the LHS
[/quote

Submitted by geanowells on Fri, 12/11/2009 - 12:42 Permalink

Based on this info;

(HPO4)2-is the weaker acid

(PO4)3- is the conjugate base

HCOOH is the weaker conjugate base

(HPO4)2- is the stronger conjugate base

For an acid, the higher the Ka value, the stronger the acid

This is not your example
H3PO4 + H2O ---> H3O+  +  H2PO4-
acid        bases     conj.acid   conj.base

HA   + H2O ---> H3O+  +  A-   
If Ka = 1 x 10-5, 'Ka' for the reverse reaction will be 1/(1 x 10-5) = 1 x 105. The larger this value, the stronger the conjugate base as it is more likely to remove H3O+ ions and move the equilibrium to the LHS

Submitted by geanowells on Fri, 12/11/2009 - 15:13 Permalink