A 100 mL sample of .18 M HClO4 is titrated with .27 M LiOH. Determine the pH of
ID: 830278 • Letter: A
Question
A 100 mL sample of .18 M HClO4 is titrated with .27 M LiOH. Determine the pH of the solution before the addition of any LiOH.
a) A 100 mL sample of .18 M HClO4 is titrated with .27 M LiOH. determine the pH of the solution after the addition of 30 mL of LiOH.
b) A 100 mL sample of .18 M HClO4 is titrated with .27 M LiOH. Determine the pH of the solution after the addition of 50.0 mL of LiOH.
c) A 100 mL sample of .18 M HClO4 is titrated with .27 M LiOH. Determine the pH of the solution after the addition of 66.67 mL of LiOH (this is the equivalance point).
d) A 100 mL sample of .18 M HClO4 is titrated with .27 M LiOH. Determine the pH of the solution after the addition of 75 mL of LiOH
e) Make a titration curve of pH vs. Volume of Base Added for this titration.
Explanation / Answer
You will need to use limiting reagent to answer this question. First thing to do is to set up the equation:
HClO4 + LiOH <---> LiClO4 + H20.
In order to find the limiting reagent you need to find the moles of HClO4 and LiOH. To find the moles of HClO4 multiply the number of Liters by the Molarity:
.1L(.18M)=.018 moles HClO4
To find the moles of LiOH do the same but with the information given for LiOH:
.03L(.27M)=8.1x10^-3 moles LiOH or .0081moles
Since .0081 moles LiOH is smaller than .018 moles of HClO4, LiOH is the Limiting Reagent. You subtract .0081 moles from both LiOH and HClO4 leaving you with .0099 moles of HClO4 and 0 moles LiOH:
.0081moles - .0081 moles = 0 moles LiOH
.018moles - .0081 moles = .0099 moles HClO4
Next you divide the remaining number of moles (.0099) by the total Liters (.1L + .03L=.13L)
.0099mol/.13L= .076 M
this also equals the [H+], next you plug this into the equation for ph:
ph=-log[H+]
ph=-log[.076]
ph= 1.12
This is a logical answer because HClO4 is a strong acid.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.