72.33 mL of 0.137 M HCl was added to an Erlenmeyer flask and is titrated with 0.165 M NaOH. Calculate the pH when 5mL of NaOH is added after the equivalence point has been reached.
First determine L NaOH for the equivalence point.
mols excess NaOH = M NaOH x L NaOH.
New M NaOH = mols excess/total volume.
(Note: total volume = volume to reach the equivalence point + 5 mL)
That will give you the M of OH^-, then pOH = -log(OH^-) followed by
pH + pOH = pKw = 14. Solve for pH.
how do you determine the L NaOH?
MHCl x L HCl = M NaOH x L NaOH
mols HCl= molsNaOH?
yes
To calculate the pH when 5 mL of NaOH is added after the equivalence point, we need to determine the concentration of the remaining HCl and NaOH in the solution.
First, we need to find the number of moles of HCl initially added. We can use the formula:
moles = concentration x volume
moles HCl = 0.137 M x 0.07233 L
moles HCl = 0.0099 mol
Since HCl and NaOH react in a 1:1 ratio, the number of moles of NaOH required to neutralize HCl will also be 0.0099 mol.
Next, we need to find the volume of NaOH required to reach the equivalence point. We can use the formula:
volume = moles / concentration
volume NaOH = 0.0099 mol / 0.165 M
volume NaOH = 0.06 L or 60 mL
Since the equivalence point has been reached, it means that all the HCl has reacted with NaOH.
Now, we need to determine the concentration of NaOH remaining after the equivalence point. We know that a total of 60 mL of NaOH was required to reach the equivalence point, and we added an additional 5 mL after that.
Total volume of NaOH after the equivalence point = 60 mL + 5 mL = 65 mL or 0.065 L
Since the number of moles of NaOH added at the equivalence point was equal to the number of moles of HCl initially present, we can use this information to find the new concentration of NaOH:
new concentration NaOH = moles / volume
new concentration NaOH = 0.0099 mol / 0.065 L
new concentration NaOH = 0.152 M
Now that we have the concentration of NaOH, we can calculate the number of moles of NaOH in the 5 mL added after the equivalence point:
moles NaOH = concentration x volume
moles NaOH = 0.152 M x 0.005 L
moles NaOH = 0.00076 mol
Since NaOH reacts in a 1:1 ratio with HCl, this means that the number of moles of HCl that were neutralized by the additional 5 mL of NaOH is also 0.00076 mol.
To calculate the resulting concentration of HCl:
moles HCl remaining = moles HCl initially - moles NaOH added
moles HCl remaining = 0.0099 mol - 0.00076 mol
moles HCl remaining = 0.00914 mol
To find the final volume of the solution:
final volume = initial volume + volume added
final volume = 72.33 mL + 5 mL
final volume = 77.33 mL or 0.07733 L
Finally, we can calculate the final concentration of HCl:
final concentration HCl = moles / volume
final concentration HCl = 0.00914 mol / 0.07733 L
final concentration HCl = 0.118 M
To calculate the pH, we can use the formula for the acidity constant (Ka) of HCl:
Ka = [H+][Cl-] / [HCl]
Since HCl ionizes completely in water, its concentration of [HCl] is equal to the initial concentration of 0.118 M. The concentration of [H+] is also equal to this value.
Taking the negative logarithm of the concentration of H+:
pH = -log[H+]
pH = -log(0.118)
pH = 0.93 (rounded to two decimal places)
Therefore, the pH when 5 mL of NaOH is added after the equivalence point is approximately 0.93.