The increase in the boiling point of a solvent is a colligative property.
That means that the increase in the boling point will be related to the number of particles (molecules or ions) present in the solution.
The higher the number of particles (molecules or ions) the higher the increase in the boiling point.
All the aqueous solutions presented are electrolytes, i.e. the solutes are ionic compounds.
Then, you have to compare the number of ions that you have in each solution.
A) 1.0 M KCl ---> 1.0 M K+ + 1.0 MCl- = 2 moles of particles / liter
B) 1.0 M CaCl2 --> 1.0M Ca(2+) + 1.0M * 2 Cl (-) = 3 moles of particle / liter
C) 2.0M KCl ---> 2.0 M K+ + 2.0 M Cl- = 4 moles of particle / liter
D) 2.0 M CaCl2 ----> 2.0 M Ca (2+) + 2.0M * 2 Cl (-) = 6 moles of particle / liter.
Then, the solution 2.0M CaCl2(aq) has the highest increase in the boiling point.
Answer: option D) 2.0 M Ca Cl2(aq)
Answer:
c = 0.0432moldm ^−3
Explanation:
The first step would be to find the molar ratio in the reaction. Now generally, one can simplify strong acid-strong base reaction by saying:
Acid+Base ->Salt+ Water
According to the reaction equation:
CH3COO- + H+ → CH3COOH
initial 0.25 0.15
change - 0.025 + 0.025
Equ (0.25-0.025) (0.15 + 0.025)
first, we have to get moles acetate and moles acetic acid:
moles of acetate = 0.25 - 0.025 = 0.225 moles
∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M
moles of acetic acid = 0.15 + 0.025 = 0.175 moles
∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M
Pka = -㏒ Ka
= -㏒ 1.8 x 10^-5
= 4.74
from H-H equation we can get the PH value:
PH = Pka + ㏒ [acetate / acetic acid]
PH = 4.74 + ㏒[0.225/0.175]
∴ PH = 4.8
