[xxOnixx]

21.60 g of sodium benzoate, NaC₆H₅COO 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
C₆H₅COO^- + H₂O  C₆H₅COOH + 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.
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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.
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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.

[Borek]

Calculate the percent ionization of the benzoate ion

What is the definition of the percent ionization?

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.

Nope, correct answer is not 4.20. Wouldn't hurt to know what you mean by x and why you tried to solve this way. In general, this is better done treating the solution as a buffer.

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.

Reaction between H⁺ and OH^- is not what dominates the solution. There is another,much more important reaction taking place. Again, this is more of a buffer problem.

http://www.chembuddy.com/?left=buffers&right=buffers

http://www.chembuddy.com/?left=buffers&right=composition-calculation

[Arkcon]

I don't see how this is a titration, but we can get to that later.

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.

I would have used Henerson-Hasselbach instead, its simpler, even if your quadratic would work (I truly don't know one way or another)  You can always used H-H to check your results.