Determining pH, [OH-], and [H30+]????

Question

How do i determine those things if i am given: .00135 M NAOH

Please also include the steps to do any problem of this type, so i can tell how to do others on my test tomorrow!!!

Answer 1

Was added 300 mL of a 0.5 M solution of NaOH that was allowed to react at room ...... determine if the switch count has reached its maximum count, if not. MRX

Answer 2

First, the easy one to answer directly- You can rewrite the concentration of NaOH as [OH-] = 1.35 X 10^-3 M.

This means that the pOH = - log [OH^-] is - log [1.35 X 10^-3 M] or approximately 2.87.

This means your pH = 14 -pOH, or
pH = 14- 2.87 = 11.13

Sometimes, you will be given the [H30+] = [H+] concentration instead, find the pH, then the pOH, then the [OH-].

so just learn three equations,

pH + pOH = 14

pH = - log [H+]
pOH = - log [OH-]

Good luck!

Answer 3

NaOH is a strong electrolyte, so it dissociates completely into Na+ and OH-
NaOH ---------> Na+ + OH-
by the stoichiometry of the reaction, u can see that the concentration of NaOH and OH- is same.
=> [NaOH] = [OH-] = 0.00135M
By the definition of pH,
pH = -log [H+]....thats the negative log of conc. of H+ in the solution.
=> pH = -log (0.00135)
u need a log table or log calculator to find this.

To calculate, the conc of [OH-], u know that the ionic product of water remains constant at a particular temperature,
=> [H+] x [OH-] = 10^(-14) (10 raised to power minus 14, which is ionic product of water at 25deg.C)
as u know [H+], u get the answer.
alternatively, u can use the following formula
(derive this urself)
14 - pH = pOH
where pOH is the negative log of [OH-] in solution.
when u know pOH, u can find [OH-] by taking antilog.
Remember one thing in such questions, if ur given a reaction, n ur asked to find the pH, then pH will be determined only by the things left in the solution AFTER THE REACTION IS COMPLETE. So, u find the pH of [H+] in the solution left after the reaction is complete.

Answer 4

The pH is equal to:

[OH-]*[H3O+]=10^-14

You know that the concentration of OH- is .00135 or 1.35*10^-3

Plug this in and you get (1.35*10^-3)*[H30+]=10^-14
or
.74*10^-11

This is the concentration of H30+ ions

Take the negative log of this to figure out the pH

It works out to 11.13

Answer 5

first calcullate pOH

pOH = log 1/[OH-]= log 1/0.00135=2.87

after this use formula pH+pOH =14

and pH= 14-pOH= 14-2.87= 11.13

and pH = log 1/[H+]=10^(-11.13)=7.4*10^(-12)

Answer 6

NaOH is a strong base

NaOH >> Na+ + OH-

[OH-] = 0.00135 M

pOH = - log [OH-] = - log ( 0.00135 ) = 2.87

pH = 14 - pOH = 11.13