21.60 g of sodium benzoate, NaC

_{6}H

_{5}COO is dissolved in enough water to make 0.750 L of solution. The K

_{a} for benzoic acid is 6.3x10

^{-5}b. Calculate the percent ionization of the benzoate ion.

That's one of the parts I'm struggling with. I wrote the chemical equation for the benzoate ion in water

C

_{6}H

_{5}COO

^{-} + H

_{2}O

C

_{6}H

_{5}COOH + OH

^{-}And found the Kb value to be 1.6x10

^{-10} via (K

_{a})(K

_{b}) = K

_{w}. That's about as far as I got.

I did find the pH, which was 8.76 through using the K

_{b} value and then solving for OH

^{-} concentration.

~~~~~~~~~~~~~~~~~~~

Calculate the pH of a solution made by adding 6.100g of benzoic acid to 250.0 mL of the sodium benzoate solution. Assume no volume change.

I ended up doing 1.6x10

^{-10} = [(0.200M + x)(x)]/0.200M. Using quadratic equation, I found x = 1.6x10

^{-10}, which was the OH

^{-} concentration and then I found that to have a pH of 4.20. But I wonder if I did it right.

~~~~~~~~~~~~~~~~~~~~

Calculate the pH of a solution made by adding 100.0mL of a 0.400M solution of HCl to a 250.0 mL of original sodium benzoate solution. Assume volumes are additive.

From calculating the pH, I found the hydroxide concentration to be 5.7x10

^{-10} Molar and then, given 250.0mL of the stuff, I got 1.4x10

^{-6} moles of OH

^{-}. Since HCl dissociates totally, from the things given to me in the problem, I got 0.04 moles of HCl in solution, implying 0.04 moles H

^{+} in solution.

So from 0.04 moles H

^{+} and basically a negligible amount of OH

^{-} in comparison, I got 0.114 M of H

^{+}pH is 0.94??? That sounds incredibly low and I feel as though there was an error in my calculations.