- 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

**-- mrcodedude**