Can someone help me with this question please?
2 answers:
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Answer:
Explanation:
Given that:
Concentration of = 0.105 M
Volume of = 20.0 mL
Concentration of = 0.125 M
The chemical reaction can be expressed as:
Using the ICE Table to determine the equilibrium concentrations.
I 0.105 0 0
C -x +x +x
E 0.105 - x x x
Recall that the ka for
Then;
By solving the above mathematical expression;
x = 0.00137 M
pH = 2.86
Hence, the initial pH = 2.86
b) To determine the volume of the added base needed to reach the equivalence point by using the formula:
Thus, the volume of the added base needed to reach the equivalence point = 16.8 mL
c) when pH of 5.0 mL of the base is added.
The Initial moles of molarity × volume
number of moles of 5.0 NaOH = molarity × volume
number of moles of 5.0 NaOH =
After reacting with 5.0 mL NaOH, the number of moles is as follows:
Initial moles
0 0
F(moles) 0
The pH of the solution is then calculated as follows:
Recall that:
pKa for
Then; we replace the concentration with the number of moles since the volume of acid and base are equal
∴
pH = 4.37
Thus, the pH of the solution after the addition of 5.0 mL of NaOH = 4.37
d)
We need to understand that the pH at 1/2 of the equivalence point is equal to the concentration of the base and the acid.
Therefore;
pH = pKa = 4.74
e) pH at the equivalence point.
Here, the pH of the solution is the result of the reaction in the with
The total volume(V) of the solution = V(acid) + V(of the base added to reach equivalence point)
The total volume(V) of the solution = 20.0 mL + 16.8 mL
The total volume(V) of the solution = 36.8 mL
Concentration of = moles/volume
=
= 0.0571 M
Now, using the ICE table to determine the concentration of ;
I 0.0571 0 0
C -x +x +x
E 0.0571 - x x x
Recall that the Ka for =
Hence, the pH of the solution at equivalence point = 8.75
f) The pH after 5.09 mL base is added beyond (E) point.
Before 0.0021 0.002725 0
After 0 0.000625 0.0021
From above; we can determine the concentration of by using the following method:
pH = 12.17
Finally, the pH of the solution after adding 5.0 mL of NaOH beyond (E) point = 12.17