Hydroxide ions (OH-) greatly increases in number and or concentration, number within a given volume, when a base is dissolved in water. This explanation proposes the Arhennius definition of a base.
The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (
for HF is
.)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = 

Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)


Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
![[H^+]=x=0.01346 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.01346%20M)
The pH of the solution is ;
![pH=-\log[H^+]=-\log[0.01346 M]=1.87](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.01346%20M%5D%3D1.87)
The pH of an 0.280 M HF solution is 1.87.
I believe it is element As, for arsenic.
<h3><u>Answer</u>;</h3>
= 226 Liters of oxygen
<h3><u>Explanation</u>;</h3>
We use the equation;
LiClO4 (s) → 2O2 (g) + LiCl, to get the moles of oxygen;
Moles of LiClO4;
(500 g LiClO4) / (106.3916 g LiClO4/mol)
= 4.6996 moles
Moles of oxygen;
But, for every 1 mol LiClO4, two moles of O2 are produced;
= 9.3992 moles of Oxygen
V = nRT / P
= (9.3992 mol) x (8.3144621 L kPa/K mol) x (21 + 273) K / (101.5 kPa)
= 226 L of oxygen