4. In a plant, sulfuric acid is purchased with a concentration of 95%. The specific gravity of the 95% sulfuric acid is 1.8358. A solution of 40 % sulfuric acid is required in the...
4. In a plant, sulfuric acid is purchased with a concentration of 95%. The specific gravity of the 95% sulfuric acid is 1.8358. A solution of 40 % sulfuric acid is required in the digester. The required volume of 40% acid solution is 1500 liters and its specific gravity is 1.3070. . The procedure requires the complete addition of water before the acid. What is the required volume of water and acid to make the 40% solution.
The mass conservation law will help us in this problem. Denote `rho_1` the known density of 95% sulfuric acid, `rho_2` the known density of 40% sulfuric acid (the given numbers are in `g/(cm^3)` ), `rho_w=1g/(cm^3)` the (known) density of pure water. Let the unknown volumes of 95% acid and water be `V_a` and `V_w,` and the known volume of resulting solution `V_2.`
Then the mass before the mixing is `V_w rho_w + V_a rho_1` and the mass after the mixing is `V_2 rho_2,` and they must be equal:
`V_w rho_w + V_a rho_1 =V_2 rho_2.`
The mass of pure sulfuric acid also remains the same, thus
`0.95 V_a rho_1 = 0.40 V_2 rho_2.`
From this equation we can find the required volume of 95% acid: `V_a = 0.40/0.95 *V_2 *rho_2/rho_1.`
Substitute it into the first equation and obtain
`V_w rho_w +0.40/0.95 V_2 rho_2 =V_2 rho_2, or V_w = V_2rho_2/rho_w (1 - 0.40/0.95).`
Numerically the answers are
`V_a = 0.40/0.95 *1500*1.3070/1.8358 approx 450` (liters of 95% acid)
`V_w = 1500*1.3070* (1 - 0.40/0.95) approx 1135` (liters of water).
Note that the volume is not preserved.