Answer:
![K=\frac{[CaO][CH_{4}][H_{2}O ]^{2} }{[CaCO_{2}][H_{2}]^4 }](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BCaO%5D%5BCH_%7B4%7D%5D%5BH_%7B2%7DO%20%5D%5E%7B2%7D%20%20%7D%7B%5BCaCO_%7B2%7D%5D%5BH_%7B2%7D%5D%5E4%20%20%7D)
Explanation:
The equilibrium expression is the K value equal to the product of the concentrations of the products over the product of the concentrations of the reactants. If there is a coefficient in front of the compound, raise the molecule to that power.
Since K is big, more product is expected. This is because of mathematic principles. A large numerator with a small denominator will produce a large number.
1. b
2.a
3.c
I think . hope I get it right
Reactivity is the name your looking for I believe.
The HCl added = 1.25 moles
and the moles of Na2HPO4 = 1 mole
Now when acid is added in the given solution of Na2HPO4
One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4
Na2HPO4 + H+ ---> NaH2PO4
Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution
So this will result into formation of a buffer of phosphoric acid and NaH2PO4
NaH2PO4 + H+ ---> H3PO4
pKa of H3PO4 = 2.1
so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58
so the pH will be in between 2.1 to 7.2
