What is the pH, pOH and [OH-] of 0.06 mol/L HI(aq)?
HI ==> H^+ + I^-
HI is a strong acid and ionizes 100%.
If (HI) = 0.06 M, then (H^+) must be 0.06 M.
pH = -log(H^+).
pOH = 14 - pH.
pOH = -log(OH^-)
Post your work if you get stuck.
Strychnine, C21H22N2)2(aq) is a weak base but a powerful poison. Calculate the pH of a 0.001 mol/L solution of strychnine. The Kb of strychnine is 1.0 x 10^-6
DrBob222, I am planning on obtaining a pHD in about nine years:P Can you give me some advice on how to be that 1% of the population who has a pHD?
Start now!
Let's call strychnine SN.
SN + HOH ==> SNH^+ + OH^-
Kb = (SNH^+)(OH^-)/(SN)
(SNH^+) = x
(OH^-) = x
(SN) = 0.001 - x
Plug into Kb and solve for x = (OH^-)
pOH = - log(OH^-)
and pH = 14 - pOH
HI ==> H^+ + I^-
HI is a strong acid and ionizes 100%.
If (HI) = 0.06 M, then (H^+) must be 0.06 M.
pH = -log(H^+).
pOH = 14 - pH.
pOH = -log(OH^-)
Post your work if you get stuck.
pH = -log(o.o6 M) = 1.22
pOH = 14 - pH
= 14 - 1.22
= 12.78
pOH = -log(OH-)
What is OH-?
I assume you have 12.78 for pOH but don't know how to get (OH^-).
You have it set up ok.
12.78 = -log(OH^-)
enter 12.78 on your calculator.
Change it to a - sign (or enter -12.78 at first). Then look for the 10x button on your calculator. Hit it and you should see 1.6666 if you carry all the calculation through from 0.06 M without rounding anything. If you round to 12.78, then change to -12.78, then hit 10x button, you get 1.6595 which rounds to 1.66 to 2 s.f.
To find the pH, pOH, and [OH-] of a solution of HI(aq) with a concentration of 0.06 mol/L, we need to understand the chemical properties of HI(aq).
HI is a strong acid that dissociates completely in water to form H+ and I- ions:
HI(aq) → H+(aq) + I-(aq)
Since HI is a strong acid, we can assume it is a strong electrolyte and that it completely ionizes in water. Therefore, the concentration of H+ ions in the solution will be equal to the initial concentration of HI.
Given that the concentration of HI is 0.06 mol/L, the concentration of H+ ions will also be 0.06 mol/L.
pH is a measure of the acidity of a solution and is defined as the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log[H+]
So, in this case, pH = -log(0.06).
To calculate the pH using a scientific calculator, take the negative logarithm of 0.06:
pH = -log(0.06) ≈ 1.22
Next, we can use the fact that in any aqueous solution, the pH and pOH values always add up to 14:
pH + pOH = 14
So, to find the pOH, we subtract the pH from 14:
pOH = 14 - pH ≈ 14 - 1.22 ≈ 12.78
Finally, we can find the hydroxide ion concentration ([OH-]) using the equation:
[OH-] = 10^(-pOH)
Substituting the calculated value of pOH, we get:
[OH-] = 10^(-12.78)
Using a scientific calculator, evaluate 10^(-12.78) to find the final answer for the hydroxide ion concentration.