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.
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To calculate the pH when 5 mL of NaOH is added after the equivalence point, we need to determine the remaining concentration of HCl in the flask and then calculate the pH based on the resulting solution.
Here are the steps to solve this problem:
Step 1: Determine the number of moles of HCl initially present in the Erlenmeyer flask before the titration.
The initial number of moles of HCl (n1) can be calculated using the formula:
n1 = C1 * V1
where:
C1 = concentration of HCl = 0.137 M (given)
V1 = volume of HCl solution added = 72.33 mL = 0.07233 L (converted from mL to L)
n1 = 0.137 M * 0.07233 L
n1 = 0.00992861 moles
So, the initial number of moles of HCl is 0.00992861 moles.
Step 2: Determine the number of moles of NaOH used to reach the equivalence point.
Since HCl and NaOH react in a 1:1 ratio (according to the balanced chemical equation: HCl + NaOH → NaCl + H2O), the number of moles of NaOH used to reach the equivalence point (n2) is also 0.00992861 moles.
Step 3: Determine the number of moles of NaOH added after the equivalence point.
The number of moles of NaOH added after the equivalence point (n3) can be calculated by subtracting the moles of NaOH used to reach the equivalence point from the total moles of NaOH added after the equivalence point.
n3 = moles of NaOH added after the equivalence point - moles of NaOH used to reach the equivalence point
n3 = (0.165 M * 0.005 L) - 0.00992861 moles
n3 = 0.000825 moles - 0.00992861 moles
n3 = -0.00910361 moles
Note: The result is negative because we have subtracted more moles than were present in the Erlenmeyer flask, which means the excess NaOH has reacted with all the HCl.
Step 4: Calculate the concentration of HCl remaining in the solution.
To determine the concentration of HCl (Cfinal) after the excess NaOH is added, use the formula:
Cfinal = n3 / V2
where:
n3 = number of moles of HCl remaining in the solution = -0.00910361 moles (calculated in step 3)
V2 = final volume of the solution = initial volume of HCl + volume of NaOH added after the equivalence point
= 0.07233 L + 0.005 L = 0.07733 L
Cfinal = -0.00910361 moles / 0.07733 L
Cfinal = -0.1177 M
The negative sign indicates that the remaining HCl is in excess and has an unknown concentration.
Step 5: Calculate the pH of the resulting solution.
To calculate the pH, we need to use the equation that relates pH with the concentration of H3O+ ions:
pH = -log[H3O+]
Since we know the concentration of HCl (in excess) remaining in the solution as -0.1177 M, we can calculate the concentration of H3O+ ions using the relationship [H3O+] = -Cfinal (given that the HCl is a strong acid and fully dissociates).
[H3O+] = 0.1177 M
Now, take the negative logarithm to find the pH:
pH = -log(0.1177)
pH = 0.93 (rounded to two decimal places)
Therefore, the pH of the resulting solution after adding 5 mL of NaOH after the equivalence point is approximately 0.93.