Hey, could anyone help explain how to "find the amount of 0.01 M NaOH needed to be added to 100mL of 0.02 M of HCl to raise the pH to 11"
Answer is 233.33 mL
Doing a question a little out of the ordinary the first things to always do is do what you would do if the question was worded as a simpler example. For instance, for this question the more standard questions are: given 25.0 mL of 0.01 M NaOH is added to 100 mL of 0.02 M HCl, find the resulting pH.
In this question we would start with writing a chemical equation and in the question that's a variation of this, it's still the same approach you will just be solving for a different variable than pH:
NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l)
Next we find moles of substances normally
c(NaOH) = 0.01 M
c(HCl) = 0.02 M, v(HCl) = 0.100 L
n(HCl) = cv = (0.02)(0.100) = 0.002 moles
Now this is where the question starts becoming different and that's when we should now start thinking about how to solve using pH = 11, but having these other things written down first makes this more easier and gives us a guide on what to do
pH = 11
pH + pOH = 14
therefore, pOH = 3 by subbing in pH = 11 to above eqn
c(OH-) excess = 10^(-pOH)
c(OH) excess = 10^(-3) = 0.001 M
The OH- is excess here because the pH is 11 indicated that more base was present
n(OH-) excess = n(OH-) initial - n(OH-) reacted with HCl
n(OH-) reacted with HCl = n(HCl)
n(OH-) reacted with HCl = 0.002 moles
n(OH-) excess = cv = (0.001)(x + 0.100)
where x is the volume of NaOH initial solution
n(OH-) initial = cv = (0.01)(x)
Sub those into the expression above
n(OH-) excess = n(OH-) initial - n(OH-) reacted with HCl
(0.001)(x+0.100) = (0.01)(x) - 0.002
Expanding gives:
0.001x + 0.0001 = 0.01x - 0.002
Collect like terms:
0.001x - 0.01x = -0.002 - 0.0001
-0.009x = -0.0021
x = 0.23333... L
x = 233.33 mL (5sf)
