I think that it is half are required to react with 40 g of HF.
No - a precipitation will occur though. Potassium nitrate is soluble in water, so the potassium and nitrate ions will remain spectator ions and stay in solution. Lead (II) hydroxide is not soluble, and will precipitate out of solution to form a solid product.
Molarity of acid=2.5M
pH=5.1.
ka=?
Now
We need to write an eqn to show the dissociation of the acid
HA + H2O === H3O+ + A-
Writing The Equilibrium(Or Acid dissociation constant) of this reaction
Ka =[H3O+] {A-]/ {HA].
The concept behind this is
concentration of Products divided by those of reactants. Water is not written because its a pure liquid and does not affect the Equilibrium constant.
Now If you have any Idea on ICE tables..
You'd know that the concentration of acid will decrease by 2.5-x
Whilst the products...will increase by x each
Note: This is when the ratio of their Moles are in 1:1
ka= x.x/2.5-x
Since the Moles of A- and H3O+ are in 1:1... Their concentrations at equilibrium will be the same
so
Ka= x²/2.5-x
Now what is x??
x is the Hydrozonium ion concentration.
we can get it from the pH formula
pH= -log (H3O+)
Making H3O+ subject by applying Logarithm Rules
H3O+ = 10^-ph
x=10^-5.1
=7.94x10^-6.
Now back to Ka
Ka= x²/2.5-x
Ka= (7.94x10^-6)²/2.5-(7.94x10^-6)
Ka= (7.94x10^-6)²/2.4999
Ka= 2.52x10^-11.
Was a Fun One
This is an example of displacement reaction
NaCl + AgF-NaF + AgCI
Answer:
33
Explanation:
as the atomic number states the number of protons in each element hope this helps :)