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
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?
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.
To find the pH of the solution after adding 50.0 mL of NaOH, we need to first determine the number of moles of NaOH added.
To do this, we can use the formula:
Number of moles = concentration × volume
Given that the concentration of NaOH is 1.00M and the volume added is 50.0 mL, we can calculate the number of moles of NaOH:
Number of moles = 1.00M × 0.0500L (since 50.0 mL = 0.0500 L)
Number of moles = 0.0500 moles
Since HCl and NaOH react in a 1:1 ratio according to the balanced chemical equation, the number of moles of HCl also present in the solution is 0.0500 moles.
Now, to find the pH of the solution, we can use the equation:
pH = -log[H+]
Since HCl is a strong acid, it fully dissociates in water to produce H+ ions. Therefore, the concentration of H+ ions is equal to the initial concentration of HCl, which is 1.00M.
Using the equation, we can determine the pH:
pH = -log(1.00M)
pH = 0
Therefore, after adding 50.0 mL of NaOH, the pH of the solution is 0 (strongly acidic).
To find the pH of the solution after adding 55.0 mL of NaOH, you can repeat the calculations using the new volume of NaOH added.