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.
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?
1 answer
2 answers