A particular vinegar is found to contain 5.8% acetic acid,CH3COOH , by mass.

What mass of this vinegar should be diluted with water to produce 0.750L of a solution with pH= 4.54?

I don't know if this problem is flawed or if you just didn't provide all of the information in the problem. First, no Ka is given for acetic acid. You will need to look that up in your text or notes. Second, the problem specifies a %w/w but then asks for the MASS to be measured to produce a VOLUME of 0.750 L. No density is given for the 5.8% stuff and no density is given for the 0.750 L; however, if you stick with the mass to weigh and assume the dilute solute is 1.0 g/mL (probably no error made here), you should be ok. A third problem is that the (CH3COOH) will be in units of M (since pH is in units of M) but you are asked to WEIGH the original solution.

CH3COOH ==> CH3COO- + H+
Ka = (H^+)(CH3COO^-)/(CH3COOH)
You want pH 4.54, convert that to H^+.
Set up an ICE chart and plug into the Ka expression above. Solve for (CH3COOH). We will call the diluted solution (2).
The original solution we will call (1). It is 5.8% w/w which means 5.8g/molar mass CH3COOH = 5.8/60 moles/100 g. That works out to be about 0.09667 moles/100 g or 0.0009667 moles/g soln but you need to confirm this. Then plug into the dilution equation as
L(2) x (CH3COOH)(2) = grams(1) x (moles/g)(1). You have all but grams(1). Solve for that. Take this will a grain of salt; I've never mixed units like this.

To find out the mass of vinegar needed to produce the desired solution, we will need to calculate the concentration of acetic acid in the solution, and then use that information to determine the mass.

Step 1: Calculate the concentration of acetic acid in the solution.
The pH of a solution can be used to determine the concentration of H+ ions present. In this case, the pH of the solution is 4.54, which means that the concentration of H+ ions is 10^(-4.54) M.

Since acetic acid is a weak acid, it partially dissociates into H+ and CH3COO- ions in water. The dissociation reaction can be represented as:
CH3COOH ⇌ H+ + CH3COO-

The concentration of acetic acid can be determined based on the concentration of H+ ions. Since the acid is monoprotic, the concentration of acetic acid is equal to the concentration of H+ ions.

Therefore, the concentration of acetic acid in the solution is 10^(-4.54) M.

Step 2: Calculate the number of moles of acetic acid needed.
The number of moles of acetic acid can be calculated using the formula:
moles = concentration × volume

In this case, the volume of the solution is given as 0.750 L. Therefore, the number of moles of acetic acid needed is:
moles = 10^(-4.54) M × 0.750 L

Step 3: Calculate the mass of vinegar needed.
To calculate the mass of vinegar needed, we need to use the percent by mass of acetic acid in the vinegar. The mass percent of acetic acid is given as 5.8%.

The mass of acetic acid present in the solution is calculated using the formula:
mass = percent mass × total mass

In this case, we want to determine the mass of vinegar needed. Let's assume the total mass of the vinegar solution is m. Therefore, the mass of acetic acid can be calculated as:
mass of acetic acid = 5.8% × m/100

We know that the moles of acetic acid in the solution is given by:
moles = mass of acetic acid / molar mass of acetic acid

We can rearrange this equation to solve for the mass of acetic acid:
mass of acetic acid = moles × molar mass of acetic acid

Since the molar mass of acetic acid is 60.05 g/mol, we can substitute these values into the equation:
5.8% × m/100 = moles × 60.05 g/mol

Step 4: Substitute the calculated values to find the mass of vinegar needed.
Substituting the calculated values from Step 1 and Step 2 into the equation:
5.8% × m/100 = (10^(-4.54) M × 0.750 L) × 60.05 g/mol

Simplifying the equation:
m = (10^(-4.54) M × 0.750 L × 60.05 g/mol × 100)/5.8%

Calculating the value:
m ≈ 0.156 kg

Therefore, approximately 0.156 kg of vinegar should be diluted with water to produce 0.750 L of a solution with a pH of 4.54.

To solve this problem, we need to find the mass of vinegar required to produce a certain volume of a solution with a specific pH.

First, let's understand the problem. The given vinegar contains 5.8% acetic acid, CH3COOH, by mass. We need to dilute this vinegar with water to produce a solution with a pH of 4.54. pH is a measure of acidity, and a lower pH value indicates a higher acidity level.

Now, let's break down the problem into steps:

Step 1: Calculate the molar concentration of acetic acid in the given vinegar.
To do this, we need to convert the mass percentage of acetic acid into molar concentration. The molar mass of acetic acid is approximately 60.05 g/mol. So, in 100 grams of vinegar, we have:

5.8 g acetic acid / 60.05 g/mol = 0.0963 mol acetic acid

This means that in 1 liter (1000 mL) of vinegar, we have 0.0963 mol acetic acid.

Step 2: Calculate the molarity of acetic acid in the given vinegar.
To do this, we need to divide the number of moles of acetic acid by the volume of the vinegar in liters:

Molarity = Moles / Volume
Molarity = 0.0963 mol acetic acid / 1 L vinegar = 0.0963 M

Step 3: Calculate the moles of acetic acid required to achieve the desired pH.
To do this, we can use the acid dissociation constant, Ka, for acetic acid. The Ka value for acetic acid is approximately 1.8 x 10^-5 at 25°C.

pH = -log[H+]
4.54 = -log[H+]
[H+] = 10^(-pH)
[H+] = 10^(-4.54)

Since acetic acid is a weak acid, we can assume that the concentration of acetic acid in the solution after dissociation is equal to [H+]:

0.0963 M = 10^(-4.54)
[H+] = 0.0963 M

So, to calculate the moles of acetic acid required to achieve the desired pH, we can use:

Moles = Molarity x Volume
Moles = 0.0963 mol/L x 0.750 L = 0.0722 mol

Step 4: Calculate the mass of vinegar required.
To calculate the mass of vinegar required, we need to multiply the moles of acetic acid by the molar mass of acetic acid:

Mass = Moles x Molar Mass
Mass = 0.0722 mol x 60.05 g/mol = 4.34 g

Therefore, you should dilute 4.34 grams of vinegar with water to produce a 0.750L solution with a pH of 4.54.