Calculate,
1,the PH of0.1M CH3COOH buffered solution with 0.2M CH3COONa
2,when 9.5M HClis added to75ml of the buffer solution in(1)
3,compaire the PH of the two questions whre added to pure water?(Ka ofCH3COOH=0.00008)
1. Use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid).
2. How much of the 9.5 M HCl is added?
3. I assume you mean compare when that many mL 0.5 M HCl is added to pure water. How much pure water.
Do #1 first then re post if you still have questions.
To calculate the pH of a buffered solution, you need to consider the equilibrium between the acid and its conjugate base. In this case, you have a solution of acetic acid (CH3COOH) and its conjugate base, sodium acetate (CH3COONa). The dissociation of acetic acid can be represented as follows:
CH3COOH ⇌ H+ + CH3COO-
1. To find the pH of the buffered solution, you need to use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log([CH3COONa]/[CH3COOH])
Here, pKa represents the negative logarithm of the acid dissociation constant (Ka) of acetic acid, and [CH3COONa] and [CH3COOH] represent the concentrations of sodium acetate and acetic acid, respectively.
Given that [CH3COOH] = 0.1 M and [CH3COONa] = 0.2 M, we can calculate the pH as follows:
pH = pKa + log([CH3COONa]/[CH3COOH])
= -log(0.00008) + log(0.2/0.1)
= 4.12 + log(2)
= 4.12 + 0.301
= 4.421
Therefore, the pH of the 0.1 M CH3COOH buffered solution with 0.2 M CH3COONa is approximately 4.421.
2. When 9.5 M HCl is added to 75 ml of the buffer solution discussed in question 1, we need to consider the reaction between HCl and the acetate ion (CH3COO-). The acetate ion acts as a buffer and helps maintain the pH. The reaction can be represented as follows:
CH3COO- + H+ ⇌ CH3COOH
To determine the resulting pH, we need to calculate the change in concentration of acetic acid and acetate ion due to the addition of HCl. Since the concentrations are given in terms of volume (75 ml), we need to convert the volume to moles using the molarity of the buffer solution.
Given that the molarity of the buffer solution is 0.1 M, the number of moles of CH3COOH and CH3COONa in 75 ml can be calculated as follows:
moles of CH3COOH = concentration × volume = 0.1 × 0.075 = 0.0075 moles
moles of CH3COONa = concentration × volume = 0.2 × 0.075 = 0.015 moles
When HCl reacts with CH3COO-, it converts it back into CH3COOH. For every mole of HCl that reacts, it consumes an equal number of moles of CH3COO-. In this case, we have excess HCl as compared to the moles of CH3COO-. Therefore, all the acetate ions will be consumed, resulting in the conversion of all CH3COO- to CH3COOH.
Since the volume of HCl added is not provided, we cannot calculate the exact moles of HCl added. However, we can assume that all the acetate ions are consumed, so the concentration of CH3COO- becomes zero. As a result, the concentration of CH3COOH will be the sum of the initial concentration of acid and the additional moles of acid produced by the reaction:
moles of CH3COOH = 0.0075 + moles of HCl
To calculate the pH after the addition of HCl, we can use the Henderson-Hasselbalch equation, which now becomes:
pH = pKa + log([CH3COONa]/[CH3COOH])
= -log(0.00008) + log(0/0.0075)
= -log(0.00008) - infinity
= -log(0.00008)
Therefore, the pH after the addition of HCl cannot be determined based on the given information, as it depends on the exact moles or volume of HCl added.
3. Comparing the pH of the two situations (buffer solution and pure water) after the addition of HCl is not possible without knowing the exact amount of HCl added or considering the volume of the solution. The pH of water is typically 7, assuming it is pure and does not contain any acidic or basic substances. When HCl is added to pure water, it will decrease the pH due to the strong acidity of HCl. However, without the specific values or volumes of HCl added in both situations, a direct comparison of the pH changes cannot be made.