I accidently posted that too early
a) 1.00mol--->1L
y--------->0.05000L
Therefore 0.05000 of NaOH was dropped into the receiving flask and neutalized.
When I calculate the amount left over 0.0500mol (I used the comparison method to get this value of pH before the titrant began)
Amount left over 0.0500mol-0.0500mol=0mol
Total volume 50.00mL(HCl)+50.00mL (NaOH)=100.00mL
Concentration of acid is now 0.00mL÷0.1000L=OM=[HCl]remaining
pH=-log [0]
pH=Math error
What was my mistake?
b)Amount----Volume
1.00mol---1L
y---------0.05500L
Therefore 0.05500mol of NaOH was dropped into the receiving flask and neutalized.
Amount left over 0.0500mol-0.05500mol=-5 x 10^-3
Total volume=50.00mL (HCl) +55.00mL (NaOH)=105.00mL
Concentration of acid is now
My answer doesn't make sense from here
What's my mistake?
Consider the case where we are titration 50.0mL of 1.00M HCl (receiving flask) with 1.00 NaOH (burette)
Find the pH of the solution:
a) After adding 50.00mL of base
I do a comparison method to determine the number of moles of NaOH
Amount----Volume
b) After adding 55.00mL of base
2 answers
a. You became more interested in the math and forgot the chemistry.
mols HCl = 0.05. mols NaOH (if that is 1.00) is 0.05
......HCl + NaOH ==> NaCl + H2O
I......0.05...0........0.....0
add..........0.05...............
C....-0.05..-0.05.....0.05..0.05
E......0.......0......0.05
so the pH is determined by the NaCl and H2O. Neither the Na^+ nor the Cl^- is hydrolyzed (they don't react with water); therefore, the pH of the solution at that point is the pH of pure water. By the way, from your first part I assume that is titrating with 1.00 M NaOH.
For part b if you will set up and ICE chart as I've done above you will see you have 5 mL of the 1.00 M NaOH too much so the (OH) = 0.005 mols/0.105 L = ? and convert that OH^- to pH.
mols HCl = 0.05. mols NaOH (if that is 1.00) is 0.05
......HCl + NaOH ==> NaCl + H2O
I......0.05...0........0.....0
add..........0.05...............
C....-0.05..-0.05.....0.05..0.05
E......0.......0......0.05
so the pH is determined by the NaCl and H2O. Neither the Na^+ nor the Cl^- is hydrolyzed (they don't react with water); therefore, the pH of the solution at that point is the pH of pure water. By the way, from your first part I assume that is titrating with 1.00 M NaOH.
For part b if you will set up and ICE chart as I've done above you will see you have 5 mL of the 1.00 M NaOH too much so the (OH) = 0.005 mols/0.105 L = ? and convert that OH^- to pH.
1 answer