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Can someone help me with this chem question ?
You recently performed a titration where you added 0.100M NaOH into
15mL of 15mL 0.100M HCl.
a) Calculate the pH of the solution in the flask at the actual equivalence
point where the moles of HCl (H3O) = the moles of NaOH (OH-
)
b) A student performs a titration and they noticed that the visual endpoint
of the titration occurred when 16.5mL of NaOH was added to the HCl.
Calculate the amount of excess OH- and the pH at the endpoint of the
titration and explain the difference between endpoint and equivalence
point.
1 Answer
- hcbiochemLv 712 months agoFavorite Answer
a) At the equivalence point, you simply have a solution of NaCl in water. The only equilibrium is:
H2O(l) <--> H+(aq) + OH-(aq)
This equilibrium is Kw = [H+][OH-] = 1.0X10^-14
Let [H+] = [OH-] = x. Then,
x^2 = 1.0X10^-14
x = [H+] = 1.0X10^-7 M
pH = 7.0
b) Excess volume of NaOH added = 1.5 mL
excess moles OH- added = 1.5X10^-4 moles OH-
Molarity OH- = 1.5X10^-4 mol / 0.0315 L = 4.76X10^-3 M
pOH = 2.32
pH = 14.00 - pOH = 11.68
The indicator used may have a rather high pH color change. Alternatively, the student may not have paying attention and missed the first indications of color change.
