If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, h
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Answer:
Option f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.
Explanation:
The pH of the buffer solution before the addition of HCl is:
The hydrochloric acid added will react with the potassium fluoride as follows:
H₃O⁺(aq) + F⁻(aq) ⇄ HF(aq) + H₂O(l)
The number of moles (η) of potassium fluoride (KF) and the HF before the addition of HCl is:
The number of moles of the HCl added is 0.408 moles. Since the number of moles of HCl is bigger thant the number of moles of KF, the moles of HCl that remains after the reaction is:
Hence, the KF is totally consumed after the reaction with HCl and thus, exceding the buffer capacity.
We can calculate the pH after the addition of HCl:
HF(aq) + H₂O(l) ⇄ F⁻(aq) + H₃O⁺(aq) (1)
The number of moles of HF after the reaction of KF with HCl is:
And the concentration of HF after the reaction of KF with HCl is is:
Now, from the equilibrium of equation (1) we have:
(2)
By solving equation (2) for x we have:
x = 0.0187
Finally, the pH after the addition of HCl is:
Therefore, the addition of HCl will exceed the buffer capacity and thus, lower the pH by several units. The correct option is f: an addition of HCl will exceed the buffer capacity. The option d is also correct since it is a consequence of the option f.
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Explanation:
It is known that charge on xenon nucleus is equal to +54e. And, charge on the proton is
equal to +e. So, radius of the nucleus is as follows.
r =
= 3.0 fm
Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.
d = r + 2.5
= (3.0 + 2.5) fm
= 5.5 fm
= (as 1 fm =
)
Therefore, electrostatic repulsive force on proton is calculated as follows.
F =
Putting the given values into the above formula as follows.
F =
=
=
= 411.2 N
or, = N
Thus, we ca conclude that N is the electric force on a proton 2.5 fm from the surface of the nucleus.