Transcendence

How to do you find the pH of a solution when you're given the concentration of it?

and how would you do this question:

You wish to prepare a solution of barium hydroxide with a pH of 13.500. What mass of the compound be required to dissolve in 500mL of water?

My teacher never really explained this :/

Oh jeez. We did these in chemistry a few weeks ago. But I didn't quite understand it myself.  :swt:

/isnohelp

hmmm if i find some time i think i'll try getting the log shit together again

ok if my memory serves me right (only doing strong bases and acids because with weak ones i'd really have to think)

the pH of a acid solution has all to do with the amount of H⁺ ions, an acid basically sheds those H⁺ ions. it doesnt love them and therefor kicks them out of the home. then a caring H₂O molecule passes by and adopts the H⁺, creating a H₃o⁺ ion.

with  strong acids it is fairly easy. 1 acid sheds 1 H⁺ ion, so 1 mole of acid gives us 1 mole of H⁺ and with that 1 mole of H₃o⁺

the first step would be to convert the amount of moles in a solution to a concentration. say i have 0,175 mole of HCl in 2L of water, this would be 0,875 mole/L

because HCl is a strong acid we can say that therefor there are 0,0875 mole/L H₃O⁺ ions in our solution.

we throw this trough a simple formula
pH = - log₁₀( H₃O⁺ )
if we fill this in we get
pH = - log₁₀( 0,0875 )
we throw this in our trusty calculator and we'll get something in the likes of
ph = 1,06


for base this is kinda the same but with different ions
now we use OH^- instead of H₃o⁺
with this you calculate
pOH = - log₁₀( OH^- )
keep in mind i said pOH, to convert that to pH you simply do
pH = 14 - pOH

2nd question of you will follow later



Keep in mind that i havent even thought about chemistry in half a year and that my memory on such things can differ a bit from reality.
if someone could confirm or reject what i said above i'd appreciate it

also i dont think it is safe for me to answer stuff about weak bases and acids as that is long ago for me, at least 2 years ago

Thanks! I got second question already (: