Question

1. The printed label on a bottle of commercial vinegar states that the acetic acid concentration is 5%. (Assume % is weight solute/volume solution)

(a) If the manufacturer had reported two significant figures in the concentration, what range of values would round to 5.0 %?

(b) Calculate the concentration in molarity of the upper and lower values from (a). The molecular weight of acetic acid is 60.05 g/mol. Pay attention to significant figures.

2. Acetic acid is a monoprotic weak acid with a pKa of 4.74. (Ka = 1.8 x 10-5)
(a) What is the pH of 5 mL of a 5.0% solution?
(b) What is the pH of the solution if you now add 45 ml of water to solution (a)?

3. How will the equivalence point volumes differ if you titrate the two solutions in question 2? Explain your answer.

Answer 1

1a) If the value is rounded off to 5.0 the actual values can be 5.00 to 5.05 .

1b) molarity =[weight /molar mass ] x 1000/volume (ml)

= [5.00x1000]/[60.05x100] = 0.8326M for the lower vlue

molarity = 5.05 x1000)/(60.05x100) = 0.8409 M for the upper value

2a) the pH of a weak acid is given by pH = 1/2 pKa - 1/2 log (concentration)

Given pKa = 4.74 and molarity (concentration) = 0.8326M (calculated in 1a)

thus pH = 1/2x4.74 - 1/2xlog(0,.8326) = 2.3278

2b) previous solution has molarity 0.8326 and now it is added with 45mL to make total volume 50 mL ..

Using M1V1 = M2V2 the molarity of diluted solution = 0.0832M

thus pH = 1/2x4.74 -1/2 log 0.0832 =1.2901

3) Since the one solution is diluted by 10 times , the equivalence point volumes also will have ten fold difference.