First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
The image is attached with the answer.
<h3>
What are Isomers ?</h3>
Isomers are compounds which have same empirical formula , same number of atoms are present but they have difference in arrangement if the atoms.
It is given that
There are four cis, trans isomers for 2-isopropyl-5-methylcyclohexanol
the cis, trans designations of the substituents are made relative to the oh group
The image is attached with this answer.
To know more about Isomers
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Answer:
45.8 mL
Explanation:
If all variables are held constant, the new volume can be found using the Boyle's Law equation. The equation looks like this:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new volume by plugging the given values into the equation and simplifying.
P₁ = 3.1 atm P₂ = 10.5 atm
V₁ = 155 mL V₂ = ? mL
P₁V₁ = P₂V₂ <----- Boyle's Law equation
(3.1 atm)(155 mL) = (10.5 atm)V₂ <----- Insert values
480.5 = (10.5 atm)V₂ <----- Multiply 3.1 and 155
45.8 = V₂ <----- Divide both sides by 10.5
Super saturated solution is formed.
<u>Explanation:</u>
Solubility is the property of any substance's capacity, that is the solute of the substance is dissolved in the given solvent to form the solution. We have three different types of solution, unsaturated, saturated and supersaturated solution.
- Unsaturated solution is a solution with lesser amount of solute than its solubility at equilibrium.
- Saturated solution is a solution with the maximum solute dissolved in the solvent.
- Super saturated solution is a solution with more solute than it is required.
The solubility of KI at 30°C is 153 g / 100 ml. Here 180 g of KI in 100 ml of water at 30°C is given, which has more solute than required, so it is super saturated solution.