A detergent solution has a pH of 11.63 at 25°C. What is the hydroxide-ion [OH-] concentration?
Is this the correct way to set it up & solve for the problem?
[OH-] = 10^-11.63
[OH-] = 2.34E-12
I wasn't sure if it was the same set up as it is for the hydronium ion.
No. The pH is 11.63, your set up calculates (H^+). I would do
14-11.63 = 2.37 = pOH, then
(OH^-) = 10^-2.37 = 4.27E-3.
Okay, thanks so much for your help!! :)
pH + pOH = pKw = 14. Solve for pOH, then pOH = -log(OH^-).
Solve for OH^-
OR you can use
pH = -log(H^+) and solve for )H^+), then
(H^+)(OH^-) = Kw = 1E-14
and solve for OH^-
Well, well, well! Looks like we have a sudsy situation here! To find the hydroxide-ion concentration, we need to use everyone's favorite - the good old pH scale! Now, pH is defined as the negative logarithm of the hydrogen-ion concentration, but lucky for us, we have the pH already. So let's turn the tables and convert it to pOH, which is the negative logarithm of the hydroxide-ion concentration.
To get the pOH, we subtract the pH from 14 (since pH + pOH = 14, they're like best buddies!). In this case, we have 14 - 11.63 = 2.37.
Now, the pOH is equal to the negative logarithm of the hydroxide-ion concentration, so we take the inverse-log (10^x) of the pOH. In this case, it's 10^(-2.37).
Drumroll, please! After crunching the numbers, we find that the hydroxide-ion concentration is approximately 0.0043 M. So, our detergent solution has a concentration of hydroxide ions that's all foamy and lathery!
To calculate the hydroxide-ion concentration ([OH-]) from the given pH value, we need to use the equation for the relationship between pH and concentration of hydroxide ions:
pOH = 14 - pH
First, let's calculate the pOH using the given pH value:
pOH = 14 - 11.63
pOH ≈ 2.37
The pOH is now known. We can then calculate the concentration of hydroxide ions ([OH-]) using the equation:
[OH-] = 10^(-pOH)
Substituting the value for pOH:
[OH-] = 10^(-2.37)
[OH-] ≈ 0.004 % M (mol/L)
Therefore, the hydroxide-ion concentration in the detergent solution is approximately 0.004 M.