Question

What is the ph of soultion with hydroxide ion concentration of 10-6m

Answer 1

\[pH = -Log[H ^{+}] \]

\[pH+pOH = 14 \]

Answer 2

\[pOH = -Log[OH ^{-}]\]

Answer 3

These are the formulas you need, you are given that the concentration of hydroxide ions in your solution is \[10^{-6} M\] so the first thing we need to do is employ the formula pOH = -Log[OH-] = 6. This means that the pOH = 6

so before I get into how to approach the answer, you must know that the pH and pOH just compare the exponents so the "p" means to take the negative log of something. so that's what we did we took the -log of the OH and we got 6.

but this isn't our answer. in-fact it's only part of the answer. we were asked to provide the pH of the solution but there is a relationship: the pH+pOH = 14

so if we know that our pOH = 6 we re-arrange our formula. \[14-pOH = pH\] to solve for the pH. and that's how you find the pH of your solution.

The info below is In case you're interested

Another thing though. why does the pH + pOH = 14?

well there's another concept, and it's called Kw (the ion product)

for water the ion product constant (to my knowledge is an equilibrium value) which implies, that it's temperature dependent. so for water at 25 degrees celsius the Kw is going to be 10^-14

let's look at this further

let's say we have water and it's neutral pH a neutral pH would be 7 because that implies that [H+] = [OH-]

this would mean that \[[H ^{+}] = [OH ^{-}] = 10^{-7} mol L ^{-1}\]

if we take the -log[10^-7] we get 7

so pH + pOH = 14 because 7+7 = 14 at 25 degrees celsius.

so the 14 comes from the ion product constant which will always be the same at that particular temperature.

Answer 4

one other thing in case you're interested.

pH only tells us the [H+] in a solution, or [OH-] pOH but it doesn't really say much about how strong the acid is. like what's the acids tendency to give up a [H+]? This idea depends on the strength of the acid that is a property intrinsic to that particular compound and thats called Ka {acid dissociation constant}

for the reaction below.

\[H _{2}SO _{4} --> HSO _{4}^{-} + H ^{+}\]

\[Ka = \frac{ [conjugate base][H+] }{ [Acid] }\]

\[K _{a} = \frac{ [HSO _{4}][H ^{+}]}{ H _{2}SO _{4} }\]

it's kind of this expression that's like ok products over reactants kind of thing. if the Ka >
1 then we have more dissociated acid. if it's less then 1 that means that the concentration of the undissociated acid is greater and that it doesn't want to give up a a proton [H+] as much.

one last thing; your \[Ka*Kb = (1x10^{-14}) at 25 \]