Finding ph, [OH-] and [H+]

Question

So I have two questions on a study guide for a test tomorrow that I just can't get. IF you could add as much work and description that would be great. step by step answers. This includes calculator inputs  Thank you so much! 

 

1) Some raspberry juice has a pH of 3.40. What is its [OH-]? 

 

2) What is the pH of a 0.02 M solution of benzoic acid? (Ka=6.3 x 10 ^ -5) 

Answer 1

1) The equation for pH = -log[H+]

If the pH = 3.40 = -log[H+]

[H+] = 10^(-3.4) = 3.98*10^-4 

 

Also, [H+]*[OH-] = 10^-14. Plug in [H+] = 3.98*10^-4 and solve for [OH-]

[OH-] = 10^-14/(3.98*10^-4) = 2.5*10^-11

 

2) Dissociation equation for benzoic acid is PhCOOH(aq) --> PhCOO-(aq) + H+(aq)

You need to compose an ICF table with each of the reagents in the equation above

Ka = [H+]*[PhCOO-]/[PhCOOH]

Since benzoic is not a strong acid, it will not dissociate completely only x amount of it will dissociate (shown in an ICF table)

Ka = x*x/(.02-x), plug in Ka which is 6.3 x 10 ^ -5 and solve for x

x^2 + 6.3*10^-3 - 0.00945 = 0 

 

Use quadratic equation, you will get x = 0.094 

x is the [H+] hence pH = 1.027

 

If you need more help understanding #2, here is a link that I found that is pretty helpful 

https://www.khanacademy.org/science/chemistry/acids-and-bases-topic/copy-of-acid-base-equilibria/v/ph-of-a-weak-acid