For the reaction of weak acid:
The Henderson-hasselbalch equation is:
![pH = pk_a+ log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%20pH%20%3D%20pk_a%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20)
The concentration ratio at 
is:
![pH = pk_1+ log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%20pH%20%3D%20pk_1%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20)
Substituting the values:
![3.2 = 4.1 + log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%203.2%20%3D%204.1%20%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20)
![log \frac{[A^{-}]}{[HA]} = - 0.9](https://tex.z-dn.net/?f=%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20%3D%20-%200.9%20%20)
![\frac{[A^{-}]}{[HA]} = 0.126](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20%3D%200.126%20%20)
Now, for calculating the pk_2 value:
![pH = pk_2+ log \frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%20pH%20%3D%20pk_2%2B%20log%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%20%20)
Thus, .
Answer:
These tests determine the solubility of the compounds formed upon adding the test solution
Explanation:
Addition of Ba(NO₃)₂ will cause a precipitate ((Ba)₂SO₄) to form in the solution of (NH₄)₂SO₄. No precipitates will form in the other unknown solutions. Thus, whether or not the solution is ammonium sulfate can be determined.
Addition of NaCl solution will cause a precipitate (AgCl) to form in the solution of AgNO₃. No precipitates will form in the other unknown solutions. Thus, whether or not the solution is silver nitrate can be determined.
If no precipitates form, then the unknown solution must be KCl.
Answer:
D) Condensation of a gas into a solid.
Explanation:
An endothermic process is a process that absorbs heat from the surroundings.
A) Boiling a liquid.
This gives off heat to the surroundings. An exothermic process.
B) Freezing a solid.
In this process, the water releases heat to the surroundings, so it is an exothermic process
C) Condensation of a gas into a liquid.
As water vapor condenses into liquid, it loses energy in the form of heat. Therefore, this process is exothermic.
D) Condensation of a gas into a solid.
This process absorbs heat from the surrounding, hence it is an endothermic proccess.
Answer:
a) in pure water:
⇒ S (25°C) Ni(OH)2 = 2.0 E-5 M
b) in 0.013 M NaOH:
⇒ S = 9.467 E-11 M
Explanation:
a) Ni(OH)2 ↔ Ni2+ + 2OH-
S S 2S
⇒ Ksp = [ Ni2+ ] * [ OH- ]² = 1.6 E-14.....Ksp value comes from the literature
⇒ Ksp = S * 2S² = 1.6 E-14
⇒ 2S³ = 1.6 E-14
⇒ S = ∛( 1.6 E-14 / 2 )
⇒ S = 2.0 E-5 M
b) NaOH ↔ Na+ + OH-
0.0130 M 0.013 0.013
Ni(OH)2 ↔ Ni2+ + 2OH-
S S 2S + 0.0130
⇒ Ksp = [ Ni2+ ] * [ OH- ]²
⇒ Ksp = S * (2S + 0.0130 )² = 1.6 E-14
Due to the small value of the Ksp it can be assumed that "s" is a very small number, so "2s" can be neglected when added to 0.0130.
⇒ 1.6 E-14 = S * ( 0.0130 )²
⇒ S = 9.467 E-11 M
Answer:
0.7158 Mol/L
Explanation:
KHC8H4O4 + NaOH → H2O + KNaC8H4O4
KHC8H4O4 molar mass = 204.2212 g/mol
0.4536 g of KHC8H4O4 = 2.221 mMol (10^-3 mol)
volume = 31.26 mL - 0.23 mL = 31.03 mL
NaOH concentration = 2.221 mMol/31.03 mL = 2.221 Mol/31.03 L = 0.7158 Mol/L
