Answer:
According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of 
and keep same equilibrium constant
Explanation:
In this buffer following equilibrium exists -
So, is involved in the above equilibrium.
When a strong base is added to this buffer, then concentration of increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of and keep same equilibrium constant.
Therefore excess amount of combines with 
to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.
Explanation:
Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.
Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.
No. of moles = 
= 
= 0.2 mol
Molarity of acetic acid is calculated as follows.
Density = 
1 g/ml = 
volume = 100 ml
Hence, molarity = 
= 
= 2 mol/l
As reaction equation for the given reaction is as follows.
So, moles of NaOH = moles of acetic acid
Let us suppose that moles of NaOH are "x".
(as 1 L = 1000 ml)
x = 20 L
Thus, we can conclude that volume of NaOH required is 20 ml.
the particles from solid to liquid start to move around faster then it was at the first state
Answer:
KOH and H₂SO₄
Explanation:
Neutralization reaction:
It is the reaction in which acid and base react with each other and produce salt and water.
For example:
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
1. Potassium hydroxide and sulfuric acid react to produce potassium sulfate salt and water.
2. Potassium hydroxide and phosphoric acid react to produce potassium phosphate and water.
H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
3. Phosphoric acid sodium hydroxide react to produce sodium phosphate and water.
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
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