Answer:
case1.
The addition of acid and base leads to a change in pH of the water when adding to deionized water due to fact that acid and bases dissociated in dissolving in water. If the H+ ion increases in the water as acid addition hikes it, it will result in decreasing the pH value. The intensity of the acid also affects the dissociation of the ions.
case2
Buffers are normally formed by weak acid and its conjugate base, and adding acid to the buffer it absorbs the H+ ions so the pH will be lower and adding base or increase of OH- conjugate base resists the pH value to increase.
Answer: The concentrations of
at equilibrium is 0.023 M
Explanation:
Moles of
= 
Volume of solution = 1 L
Initial concentration of
= 
The given balanced equilibrium reaction is,

Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :

By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of
at equilibrium is 0.023 M
To find the empirical formula you would first need to find the moles of each element:
58.8g/ 12.0g = 4.9 mol C
9.9g/ 1.0g = 9.9 mol H
31.4g/ 16.0g = 1.96 O
Then you divide by the smallest number of moles of each:
4.9/1.96 = 2.5
9.9/1.96 = 6
1.96/1.96 = 1
Since there is 2.5, you find the least number that makes each moles a whole number which is 2.
So the empirical formula is C5H12O2.
Answer: The molality of solution is 17.6 mole/kg
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
where,
n = moles of solute
= weight of solvent in kg
moles of acetone (solute) = 0.241
moles of water (solvent )= (1-0.241) = 0.759
mass of water (solvent )=
Now put all the given values in the formula of molality, we get
Therefore, the molality of solution is 17.6 mole/kg