Sorry, bit of background for clarity... or confusion. Here is the full question, and the answer supplied by the course co-ordinator in red. There was also a table supplied to show how much Ca, Cl2 and SO4 are in the Melbourne and London water profiles in ppm, and the idea was to find a way to increase Ca from 1.3 to 90ppm, and have Cl2 (molecular weight 71) and SO4 (molecular weight 96) in a 3:1 ratio. This changes one areas water profile to the other by the mixing two salts (using the known molecular weights, and with the starting values of 1.3 ppm Ca, 6.5ppm Cl2 and 0.9 SO4 already present in Melbourne water). It's a beer brewing problem.
Question:
Your water composition is the same as Melbourne, and Portersare traditionally made from “London” water. For this exercise, you want to increasethe calcium to 90 ppm (from 1.3ppm) and have a ratio of sulphate to chloride of 3:1. You haveaccess to the following salts:
CaCl2.2H2O Calcium Chloride
CaSO4.2H2O Calcium Sulphate / Gypsum
How much calcium chloride and gypsum will you add per 1000 L of water?
Answer:
Let x = ppm Ca from CaCl2.2H2O
Let y = ppm Ca from CaSO4.2H2O
You want x + y to equal 90, less the Ca in Melbourne water, so x + y = 88.7 and y = 88.7 – x
So for every bit of Ca you add from CaCl2.2H2O, you add 71/40 of Cl (2xCl, so 2 x 35.5)
So for every bit of Ca you add from CaSO4.2H2O, you add 96/40 of SO4 (1 x 32, + 4 x 16)
The ppm Cl– added = x * 71/40
The ppm SO42– added = y * 96/40
You want sulphate to chloride = 3
You already have 6.5 ppm Cl and 0.9 ppm SO4
Therefore: 3 * ([x * 71 / 40] + 6.5) = [y * 96 / 40] + 0.9
Substitute (88.7 – x) for y in above, solve the equation, and I get x = 25.1 ppm Ca as CaCl2.2H2O and y = 63.6ppm Ca as CaSO4.2H2O.
Calculate grams of salts (eg 25.1 x 147/40 for CaCl2.2H2O), and get 92.4 and 273.3ppm as CaCl2.2H2O and CaSO4.2H2O respectively. The units ppm (mg/L) are the same as g / 1000L.
While I understand the overarching principles here, I always bomb out with the maths when attempting to solve 3* ([x * 71 / 40] + 6.5) = [y * 96 / 40] + 0.9
(whereby x = 25.1 and y = 63.6 as per the feedback from the course co-ordinator)
I've been beginning with the substitution for y like this:
3* ([x * 71/40] + 6.5) = [88.7-x * 96/40] + 0.9
and trying to finish with the given x=25.1
However, I'm not sure what the first step in the order of operations is, or what to do with the x * fraction portion of the problem. For x * 71/40 I've been solving as 1.775x. Clearly that's wrong.