# how do you find pH?

Discussion in 'DAT Discussions' started by joonkimdds, Jun 2, 2008.

1. ### joonkimdds Senior Member 7+ Year Member

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I have bunch of pH questions and don't know how to solve any of these.
could you guys show me how to do at least one of these?

1) what is the pH of 0.0111 M NaOH?
2) what is the pH of 5.04 x 10^-3 M HI?
3) what is the pOH of 2.55M Ba(OH)2?
4) what are [H3O+][OH-] and pOH in a solution with a pH of 9.78?
5) what are [H3O+][OH-] and pH in a solution with a pOH of 10.48?

1) 12.045
2) 2.298
3) -0.708
4) 1.7 x 10^-10M, 6.0 x 10^-5M, 4.22
5) 2.7x10^-4M, 3.7x 10^-11M, 3.57

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Last edited: Jun 2, 2008
2. ### osimsDDS 5+ Year Member

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You have to know how to use the log to convert it to a normal number...

So for number 1 since its a base you need to find the pOH first then use 14=ph + pOH to find the ph.

So....1) what is the pH of 0.0111 M NaOH?

.0111 is equal to 1.11x10^-2
Then plus this into the equation: pOH= - log [OH-]
You will get pOH= -log [1.11x10^-2]
Then you bring the 2 down and it will be.... pOH= 2 - log 1.11
So let me just say this....if you have a log close to 10 itll be minus 1 and a log close to 1 will be very close to 0. So it will be close to 2 but not 2...around 1.8-1.9 (you have to estimate).
Now you use 14=pH-pOH to find pH....14=pH + 1.8
pH= around 12.2 or so...your answer is 12 which is close to this

3. ### Sea of ASH 2+ Year Member

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just play with the formulas and ur good to go

pH + pOH = 14

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

Useful tip:
remember that p of anything is equal to -log(of anything)

so pKa = -log(Ka) and pKb = -log(Kb)