Answer:
Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]
Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]
Explanation:
An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:
HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)
And Ka, the acid dissociation constant is:
<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />
When base:
HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)
And kb, base dissociation constant is:
<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>
Answer:
Pressure of hydrogen gas = 695.2 mmHg
Explanation:
Given:
Water temperature = 22°C
Pressure inside the tube = 715 mmHg
Find:
Pressure of hydrogen gas
Computation:
Using vapor pressure of water table
Water pressure at 22°C = 19.8 mmHg
Pressure inside the tube = Pressure of hydrogen gas + Water pressure at 22°C
715 = Pressure of hydrogen gas + 19.8
Pressure of hydrogen gas = 715 - 19.8
Pressure of hydrogen gas = 695.2 mmHg
Answer:
The K sp Value is 
Explanation:
From the question we are told that
The of 
is = 122.5 g/ mol
The mass of dissolved is 
The volume of solution is 
The number of moles of is mathematically evaluated as
Substituting values
Generally concentration is mathematically represented as
For
The dissociation reaction of is
The solubility product constant is mathematically represented as
Since there is no ionic reactant we have
![K_{sp} = [k^+] [ClO_3^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5Bk%5E%2B%5D%20%5BClO_3%5E-%5D)
Answer:
1) The bubbles will grow, and more may appear.
2)Can A will make a louder and stronger fizz than can B.
Explanation:
When you squeeze the sides of the bottle you increase the pressure pushing on the bubble, making it compress into a smaller space. This decrease in volume causes the bubble to increase in density. When the bubble increases in density, the bubble will grow and more bubbles will appear. Therefore, Changing the pressure (by squeezing the bottle) changes the volume of the bubbles. The number of bubbles doesn't change, just their size increases.
Carbonated drinks tend to lose their fizz at higher temperatures because the loss of carbon dioxide in liquids is increased as temperature is raised. This can be explained by the fact that when carbonated liquids are exposed to high temperatures, the solubility of gases in them is decreased. Hence the solubility of CO2 gas in can A at 32°C is less than the solubility of CO2 in can B at 8°C. Thus can A will tend to make a louder fizz more than can B.
Answer:
The answers are explained below
Explanation:
a)
Given: concentration of salt/base = 0.031
concentration of acid = 0.050
we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.031/0.050) = 1.59
b)
we have HSO₃⁻ + OH⁻ ------> SO₃²⁻ + H₂O
Moles i............0.05...................0.01.................0.031.....................0
Moles r...........-0.01.................-0.01................0.01........................0.01
moles f...........0.04....................0....................0.041.....................0.01
c)
we will use the first equation but substituting concentration of base as 0.031 + 10ml = 0.031 + 0.010 = 0.041
Hence, we have
PH = PK a + log[salt]/[acid] = 1.8 + log(0.041/0.050) = 1.71
d)
pOH = -log (0.01/0.510) = 1.71
pH = 14 - 1.71 = 12.29
e)
Because the buffer solution (NaHSO3-Na2SO3) can regulate pH changes. when a buffer is added to water, the first change that occurs is that the water pH becomes constant. Thus, acids or bases (alkali = bases) Additional may not have any effect on the water, as this always will stabilize immediately.
