[bpeck8]

Question 1:
Calculate the blend ratio the plant manager should use, in order to combine the vinegar sample you calculated ( which i calculated to be 2.60%) with the 8.35% batch that finished fermenting this morning to yield a final batch whose acetic acid concentration is 4.85%.


Question 2:
How many gallons of the 8.35% batch are needed to salvage 20,000 gal of the overly diluted batch?



I am not sure how to start these problems off to calculate them. Can any one help me get started and ill see if i can finish it off? Sorry i dont have any work to show, i literally have no clue on how to approach these problems.

[Borek]

Assume x g of one and y g of the other. What is the final mass of the mixture? Can you write equation showing what will mass be percentage of vinegar in the mixture? Now, assume final mass to be 100 g, and final concentration to be 4.85 - and solve for x&y.

In general first equation is not necessary, it is enough to solve the second equation for x/y ratio. But I know from experience that struggling students prefer less abstract approach

[bpeck8]

Thanks Borek for trying to help me out. I am still confused on what you are trying to tell me.

I know that a general ratio would be something like 8.35g/2.60g = 3.21g, which would mean that 3 g of morning batch roughly to 1 g of the diluted batch. I think anyways. If this has anything to do about these questions to helping me understand.

[Borek]

If you mix x g of 2.60% vinegar with y g of 8.35% vinegar, what will be the final mass of your mixture?

How much vinegar in x g of 2.60% solution?

How much vinegar in y g of 8.35% solution?

How much vinegar in total?

[bpeck8]

If you mix x g of 2.60% vinegar with y g of 8.35% vinegar, what will be the final mass of your mixture?

How much vinegar in x g of 2.60% solution?

How much vinegar in y g of 8.35% solution?

How much vinegar in total?

OK here it goes, More than likely wrong.

2.60 + 8.35 = 10.95
100 -10.95 = 89.05

89.05 = 2.60x
89.05 = x
2.60
34.25 = x

89.05 = 8.35y
89.05 = y
8.35
10.66 = y

34.25 + 10.66 = 44.91

hehe its a try, be patient and believe in me still, we'll get there.

[Borek]

If you mix x g of 2.60% vinegar with y g of 8.35% vinegar, what will be the final mass of your mixture?

2.60 + 8.35 = 10.95

Huh?

If you mix x grams of something with y grams of something, what is the final mass?

[bpeck8]

So i finally had help from a professor today. this is what he came up with for me.

2.60% (vd) + 8.50% (vc) = 4.85% 1 gal                                    vd + vc = 1 ( i think this is what you were trying to help me understand and get)
2.60 (vd) + 8.50 ( 1-vd) = 4.85                                               vc = 1 - vd
2.60vd + 8.50 - 8.50vc = 4.85                                                vc = 1 - .609
                     -5.75vd = -3.5                                                 vc = .391
                            vd = .609

vc/vd = .391/.609 ratio



(.391vc)(20,000 vd gal) = 12,841 vc gal 8.35%
(.609vd)

Can't believe i wasn't thinking like this. So simple now that i think about it.

[Borek]

vd + vc = 1 ( i think this is what you were trying to help me understand and get)

More or less.

One thing is slightly wrong in your prof approach IMHO. Volumes are not additive, so you can't assume

2.60% (vd) + 8.50% (vc) = 4.85% 1 gal                                    vd + vc = 1

are right at the same time. Final volume of mixture is a function of initial concentrations. Final differences are usually pretty small, but they do exist.

In this particular case you need to mix 620.7 mL of diluted (2.60%) acid with 379.4 mL of concentrated one (8.50) to get 1000 mL of 4.85% solution. Sum of volumes is 620.7mL+379.4mL = 1000.1mL, but there is a small volume contraction when mixing and the final volume will be exactly 1000 mL. Difference is small, but you don't know how large it is beforehand. On teh other hand, if you will use masses of the solutions instead of volumes your final result will be exact. After you know masses you have to convert them to volumes using densities taken from the density tables.