check my work please, find the concentration of H+, OH-, pH and percent ionization of the acid.?

Question

Hi, will you all check my answers and significant figures, thanks

A mass of 0.256g of benzoic acid is dissolved in enough water to obtain 100.0 mL of solution. Calculate the concentration of H+ and OH- in the solution, the pH and the percent ionization of the acid.

first I got pKa from a site for benzoic acid, it is 4.20, I got the Ka by 10^-4.20=
6.3x10^-5 = Ka

now I start out by getting the moles of benzoic acid by taking .256g and deviding by 122 which gives me .00210 mol

Now that I have the moles i get the molarity by deviding the moles by liters of volume, which is .100L. So 0.00210/.100 = 0.0210M

I use the ICE method to set up my equation which is, x^2/(0.0210-x)=6.3x10^-5 I solve for x which gives me .00111M this is also the [H+] to get [OH-] I devided 10^-14 by .00111M = 9.1x10^-12 = [OH-]

Now to get pH I take the -log of [H+] so -log [.00111M] = 2.96pH

And last % ionization = .00111/.0210=.0529 x 100 = 5.29%

Are all my values correct? also sig figs

Answer 1

Your calculations are O.K. I found a pH of 2.94 but that is an insignificant difference.

Answer 2

pH = log[H+] An acid is something that donates H+ ions. The more concentrated they are in the solution, the stronger the acid. pH of an acid = 1-5 A base is a solution with a low concentration of H+ and a higher concentration of OH- ions. pH of a base 14-9 Water is a neutral solution and has a pH of around 7.