0.250 moles of HCL and 0.100 moles of acetic acid are mixed in 0.365 moles of water. Determine the following: [CH3COOH], [CH3COO^-], pH and pOH.
Ka=1.8E-5.
Calculate the pH of 0.05M CH3COO-Na+ .
Plz show me step by step, really confused right now.
If I read this problem correctly it provides numbers that are entirely unrealistic and I don't want to tackle it. For example, 0.365 mol H2O = approx 6.6 grams. If the density is 1.0 g/mL that is about 0.0066 liters which makes the HCl about 0.250/0.0066 = approx 38 M and there is no such animal.
But these are the numbers which have been provided in the question...
If u could show me just the way to tackle it, it wud be very helpful,plz.
To solve this problem, follow these steps:
Step 1: Find the concentration of acetic acid ([CH3COOH]):
To find the concentration of acetic acid, divide the number of moles of acetic acid (0.100 moles) by the volume of water (0.365 moles):
[CH3COOH] = 0.100 moles / 0.365 L = 0.274 M
Step 2: Find the concentration of acetate ion ([CH3COO^-]):
Since acetic acid is a weak acid, it dissociates into its conjugate base, acetate ion ([CH3COO^-]). The concentration of acetate ion is equal to the concentration of acetic acid:
[CH3COO^-] = 0.274 M
Step 3: Calculate the pH:
The pH is determined by the concentration of hydrogen ions ([H+]), which can be found using the dissociation constant (Ka) of acetic acid:
Ka = [H+][CH3COO^-] / [CH3COOH]
Rearrange the equation to solve for [H+]:
[H+] = Ka * ([CH3COOH] / [CH3COO^-])
[H+] = (1.8E-5) * (0.274 / 0.274)
[H+] = 1.8E-5 M
To calculate the pH, use the equation:
pH = -log[H+]
pH = -log(1.8E-5)
pH ≈ 4.75
Step 4: Calculate the pOH:
pOH is determined by the concentration of hydroxide ions ([OH-]) in the solution, which can be found using the dissociation constant (Kw) of water:
Kw = [H+][OH-]
Since the solution is neutral, [H+] = [OH-], and Kw can be expressed as:
Kw = [H+]^2
Rearrange the equation to solve for [OH-]:
[OH-] = [H+]
To calculate the pOH, use the equation:
pOH = -log[OH-]
pOH = -log(1.8E-5)
pOH ≈ 4.75
Step 5: Calculate the pH of 0.05 M CH3COO-Na+:
To calculate the pH of the sodium acetate solution, you need to consider the hydrolysis reaction of acetate ion with water. Acetate ion reacts with water to form acetic acid and hydroxide ion.
The equilibrium expression for the hydrolysis reaction is:
CH3COO^- + H2O ⇌ CH3COOH + OH-
Since the concentration of CH3COO^- is 0.05 M, and we already know the concentration of CH3COOH and OH- (from the previous calculations), we can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([CH3COOH] / [CH3COO^-])
In this case, the pKa is the negative logarithm of the dissociation constant (Ka):
pKa = -log(1.8E-5)
pKa ≈ 4.74
Substituting the values into the equation:
pH = 4.74 + log (0.274 / 0.05)
pH ≈ 4.74 + log (5.48)
pH ≈ 4.74 + 0.74
pH ≈ 5.48