2.4 Mixing Solutions of Different Concentrations to Achieve a Target Concentration

Figure: Mixing a weaker and a stronger solution gives something in between.

This is mathematically identical to the two-ingredient price alligation — simply replace “price per unit” with “percentage concentration.” The criss-cross diagram works exactly the same way.

Worked Example 2.4.1: In what ratio should a 40% acid solution be mixed with a 60% acid solution to get a solution that is 45% acid?

Solution: C_1=40, C_2=60, M=45. Ratio = (60-45):(45-40) = 15:5 = 3:1 Mix 40% solution and 60% solution in the ratio 3 : 1.

Worked Example 2.4.2: How many litres of a 30% alcohol solution should be mixed with a 50% alcohol solution to make 20 litres of a 45% alcohol solution?

Solution: Ratio = (50-45):(45-30) = 5:15 = 1:3. Total parts = 1+3=4. So: 30% solution = 20 × (1)/(4) = 5 L, 50% solution = 20×(3)/(4) = 15 L

Worked Example 2.4.3: 300 g of a sugar solution has 40% sugar in it. How much sugar should be added to make the solution 50% sugar?

Solution: Sugar =120 g, the “rest” (water) =180 g and this quantity of water stays unchanged as we only add sugar. Let y g of sugar be added: (120+y)/(300+y) = (50)/(100) ⇒ 240+2y = 300+y ⇒ y = 60 g

Worked Example 2.4.4 (diluting instead of concentrating): What quantity of water must be added to 30 L of a solution containing 40% acid, to reduce the acid concentration to 25%?

Solution: Acid quantity =30×0.4=12 L, and this stays fixed as only water is added. Let w L of water be added: (12)/(30+w)=0.25 ⇒ 30+w=48 ⇒ w=18 L

Worked Example 2.4.5 (three solutions mixed to a target concentration — Type A weighted average applied to percentages): Three solutions containing 20%, 40% and 60% acid respectively are mixed in the ratio 1 : 1 : 2. Find the percentage of acid in the resulting mixture.

Solution: This is a direct weighted average, exactly like Section 2.2’s Type A, but with percentages instead of prices: M = (20(1)+40(1)+60(2))/(1+1+2) = (20+40+120)/(4) = (180)/(4) = 45 The resulting mixture contains 45% acid.

Worked Example 2.4.6 (edge case — adding the pure ingredient itself, not water, to raise concentration): A vessel contains 40 L of a solution that is 10% acid. How many litres of pure acid must be added to raise the concentration to 25%?

Solution: Here, unlike Examples 2.4.3/2.4.4, we are adding pure acid, not water — so it is the non-acid (water) part that stays fixed, not the acid part. Acid initially =40×0.10=4 L; water =36 L (constant, since only acid is added). Let y L of pure acid be added: (4+y)/(40+y) = 0.25 ⇒ 4+y = 10+0.25y ⇒ 0.75y = 6 ⇒ y=8 Check: Acid =4+8=12 L, total =48 L, (12)/(48)=25% ✓. 8 L of pure acid must be added.