Determine the [H3O+] of a 0.100 M solution of benzoic acid.
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Question “Determine the [H3O+] of a 0.100 M solution of benzoic acid.”
Answer
C6H5COOH = Benzoic acid. Because benzoic is a weak acid we must do a hydrolysis and an ICE to determine how much it separates into H3O+ ions.
C6H5COOH(aq) + H2O(l) <--> C6H5COO-(aq) + H3O+(aq)
Initial 0.100M / Zero 0
Change -x/+x +x
Equil. Equil.
We don’t care about H 2O, because we make an ICE table to determine equilibrium concentrations to use them in equilibrium constant expressions that only include aqueous and gaseous solutions.
Let’s now determine our equilibrium expression. The equilibrium expression is the sum of the concentrations at equilibrium of the products to the power of the coefficients and the concentrations at equilibrium of the reactants to the powers of the coefficients.
K for benzoic acids is 6.5*10 5 (these values can be found in the appendix to your textbook). We can therefore substitute the equilibrium concentrations and the K value in the equation above to get:
We can subtract 0.100M from (0.100M-x), because K for benzoic acids is so small. This doesn’t change the overall answer.
We get:
However, our ICE table shows that x is the equilibrium concentration of H O + so we have
[H3O+]= 0.00255M
If you have any questions, let me know!
Conclusion
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