Answer:
The answer to your question is pH = 1.45
Explanation:
Data
pH = ?
Volume 1 = 200 ml
[HCl] 1 = 0.025 M
Volume 2 = 150 ml
[HCl] 2 = 0.050 M
Process
1.- Calculate the number of moles of each solution
Solution 1
Molarity = moles / volume
-Solve for moles
moles = 0.025 x 0.2
result
moles = 0.005
Solution 2
moles = 0.050 x 0.15
-result
moles = 0.0075
2.- Sum up the number of moles
Total moles = 0.005 + 0.0075
= 0.0125
3.- Sum up the volume
total volume = 200 + 150
350 ml or 0.35 l
4.- Calculate the final concentration
Molarity = 0.0125 / 0.35
= 0.0357
5.- Calculate the pH
pH = -log [H⁺]
-Substitution
pH = -log[0.0357]
-Result
pH = 1.45
Lower than 7 is acid greater than 7 is a base
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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