- Jun 23, 2001
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The pH of a 0.175M solution of the hypothetical acid HOW is 5.87. Calculate the Ka for this acid.
HOW (aq) <--> (represents equilibrium) H+(aq) * OW-(aq)
Here are steps to solving the problem.
1. Write the equilibrium equation and Ka expression for the acid (already done)
2. Convert pH to [H+] ( equation is: pH = -log [H+] )
3. Determine the concentrations of all substances in the equilibrium equation.
4. Substitute into Ka expression and solve for Ka.
I got my answer to come out to be 1.04 x 10^-11, while my friend got 9.6 x 10^-12
Thanks in advance to anyone who helps![Smile :) :)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
-- mrcodedude
HOW (aq) <--> (represents equilibrium) H+(aq) * OW-(aq)
Here are steps to solving the problem.
1. Write the equilibrium equation and Ka expression for the acid (already done)
2. Convert pH to [H+] ( equation is: pH = -log [H+] )
3. Determine the concentrations of all substances in the equilibrium equation.
4. Substitute into Ka expression and solve for Ka.
I got my answer to come out to be 1.04 x 10^-11, while my friend got 9.6 x 10^-12
Thanks in advance to anyone who helps
-- mrcodedude