How do acids and bases interact in a buffer solution?

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Calcium oxide reacts with water to form calcium hydroxide, represented by the equation CaO + H2O → Ca(OH)2. In a buffer solution with HF, adding a small amount of NaOH will shift the equilibrium according to Le Chatelier's principle, resulting in an increase in pH due to the formation of more F⁻ ions. The addition of NaOH neutralizes some of the H3O⁺ ions, reducing their concentration. For the acetic acid question, dissolving 0.75 mol of acetic acid in 3.0 L of water results in a specific pH that can be calculated using the acid dissociation constant (Ka) for acetic acid. Understanding these interactions is crucial for manipulating pH in buffer solutions.
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1.) i would like to know if this is right
What is the reaction that occurs when calcium oxide is added to water?
is it : CaO + H2o --> Ca(OH)2

and the second question which i have no idea to is..

Consider the buffer equilibrium:

HF(aq) + H2O(l) ---> H30(aq) + F(negative)(aq)
the HF has high conc.
The H3O has low conc.
and the F has high conc.
Using le Chat's principle, what happens to the pH of the buffer solution when a small amount of NaOH is addeD?!
thanks a bunch u guys!
 
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help.. :(
anyone..specially the second question
 
yikes, i also have another question... please help.. :(

What pH results when .75 mol of acetic acid is dissolve din enough water to make 3.0L of solution..?
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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