Answer:
Benzoic acid is the stronger acid
Explanation:
Weak acids do not dissociate completely in the solution. They exists in equilibrium with their respective ions in the solution.
The extent of dissociation of the acid furnising hydrogen ions can be determined by using dissociation constant of acid (
).
Thus for a weak acid, HA
The is:
![K_a= \frac{[A^-][H^+]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%20%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D)
The more the , the more the acid dissociates, the more the stronger is the acid.
Also,
is defined as the negative logarithm of .
So, more the , less is the and vice versa
All can be summed up as:
The less the value of , the more the is and the more the acid dissociates and the more the stronger is the acid.
Given,
of acetic acid = 54.7
of benzoic acid = 54.2
of benzoic acid < of acetic acid
So, benzoic acid is the stronger acid.
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4 elements are classified in group 4 like titanium, zirconium, hafnium and ruthorfordium
Answer:
I am pretty sure Danny Duncan told me 69
Explanation:
niice
Answer:
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
Explanation:
Step 1: Data given
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = Shows how much the boiling point increases
⇒i = the van't Hoff factor: Says in how many particles the compound will dissociate
⇒ Since all are aqueous solutions Kb for all solutions is the same (0.512 °C/m)
⇒m = the molality
Step 2:
0.20 m glucose
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for glucose = 1
⇒ Kb = 0.512 °C/m
⇒m = 0.20 m
ΔT = 1*0.512 * 0.20
<u>ΔT = 0.1024 °C</u>
0.30 m BaCl2
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for BaCl2 = Ba^2+ + 2Cl- : i = 3
⇒ Kb = 0.512 °C/m
⇒m = 0.30 m
ΔT = 3*0.512 * 0.30
<u>ΔT = 0.4608 °C</u>
0.40 m NaCl
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for NaCl = Na+ + Cl- : i = 2
⇒ Kb = 0.512 °C/m
⇒m = 0.40 m
ΔT = 2*0.512 * 0.40
<u>ΔT = 0.4096 °C</u>
0.50 m Na2SO4.
ΔT = i*Kb*m
⇒ΔT = the boiling point elevation = TO BE DETERMINED
⇒i = the van't Hoff factor for Na2SO4 = 2Na+ + SO4^2- : i =3
⇒ Kb = 0.512 °C/m
⇒m = 0.50 m
ΔT = 3*0.512 * 0.50
<u>ΔT = 0.768 °C</u>
0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.
