Homework Help: Chemistry

Posted by Katherine on Friday, April 20, 2012 at 1:46am.

C6H5COOH(aq) + NO2-(aq) <=> C6H5COO-(aq) + HNO2(aq)

A table of ionizations is given to calculate the equilibrium constant for this overall reaction.. but when i looked onto the table the only thing given was the Ka for C6H5COOH [6.3×10^–5] and HNO2 [5.6×10^–4]

it doesn't show me the Ka for NO2- or C6H5COO- ... is there a way I can solve K without knowing the Ka for those last two..? please help! it'll be greatly appreciated!

Chemistry - DrBob222, Friday, April 20, 2012 at 9:21pm
#1.
C6H5COOH ==> C6H5COO^- + H^+ k1 = above.

#2. Reverse HNO2 ionization which is k2 = above.
NO2^- + H^+ ==> HNO2 k' = 1/k2

Add #1 to #2 and you obtain
C6H5COOH + NO2^- ==> C6H5COO^- + NO2^- and Keq for that reaction is k1*k'
Chemistry - Katherine, Saturday, April 21, 2012 at 2:38pm
Thanks DrBob!

Answer this Question

Related Questions

Chemistry, acids and bases - (a) Find pH and pOH of 1.0 M solution of sodium ...
Chemistry (conceptual question) - If you have a soln of HNO2 at equilibrium HNO2...
College Chemistry - Given the following equilibrium constants, Ka (NH4^+)=5.6*10...
Chemistry - I wrote the dissociation or ionization in water for each of the ...
Chemistry - Use the data in Table 15.3 to calculate the equilibrium constant for...
chemistry - write equilibrium constant expression for each reaction. a) The ...
chemistry - Could you help me complete these equations. I can balance them after...
chemistry... HELPP - Chemistry Equilibrium Constant PLEASE Help!? In an ...
chemistry... HELP - Equilibrium constant question!!!? Standard reduction ...
Chemistry - Write equilibrium Constant expressions (k) for the following ...

For Further Reading

First Name:
School Subject:
Answer: