Respuesta :
Answer:
(a) adding 0.050 mol of HCl
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -Weak acid-
1.0L × (0.050 mol / L) = 0.050 moles of NaF -Conjugate base-
-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-
Thus:
(a) adding 0.050 mol of HCl: The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid destroying the buffer.
(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer wouldn't be destroyed.
(c) adding 0.050 mol of NaF: The addition of conjugate base doesn't destroy the buffer
The action that destroys the buffer should be option (a) adding 0.050 mol of HCl
What is buffer?
It refers to the mixture of the weak acid also it deals with the conjugate base or it can be vice versa
We know that
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -Weak acid-
1.0L × (0.050 mol / L) = 0.050 moles of NaF -Conjugate base-
So the weak acid should be reacted with the base as NAOH and the conjugate base should be reacted with the acid as HCI
Now in the case when the 0.050 moles of HCI should be added so it generates the reaction of 0.050 moles. here the NaF should be consumed due to which it destroyed the buffer.
Therefore, the option a is correct.
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