Suppose you prepare a solution that is 30 wt% h2so4 in water and intend to confirm the concentration by comparing the specific gravity (sg) to a value obtained from the literature. you slowly added 30 g h2so4 to 70 g h2o and allowed the resulting mixture to equilibrate at 20oc. i). estimate the volumes of h2o and acid that were blended and estimate the density of the 30 wt% acid solution assuming volume additivity. ii). if the handbook sg value of the 30 wt% acid is given as 1.2185, estimate the volume of 30 wt% acid prepared as above. iii). what volume of 40 wt% acid (sg = 1.2991) must be added to 100 g of the 30 wt% solution to produce a solution that is 35 wt% h2so4?

i) V H2O = 70.13 mL ; V H2SO4 = 16.39 mL ; d 30 wt% = 1.1558 g/mL

ii) 82.07 mL Solution

iii) 76.98 mL of 40 wt% solution

Explanation:

i) According to the literature, at 20 ° C the density of water (H2O) and sulfuric acid (H2SO4) is approximately 0.9982 g / mL and 1.83 g / mL, respectively. With these values, we can know how many milliliters are the grams of each substance equals, making the following conversion (remembering that when in a division there are the same units above and below, they can cancel each other):

70 g H2O * (1 mL / 0.9982 g) = 70.13 mL

30 g H2SO4 * (1 mL / 1.83 g) = 16.39 mL

On the other hand, the density of the 30 wt% acid solution would be given by the following equation:

(1) d 30 wt% = g solution / mL solution

The total mass of the solution would be:

70 g H20 + 30 g H2SO4 = 100 g Solution

While assuming that the volumes of H2O and H2SO4 are additive, the total volume of the solution would be:

70.13 mL H2O + 16.39 mL H2SO4 = 86.52 mL Solution

Substituting these values in equation (1):

(1) d 30 wt% = 100 g / 86.52 mL = 1.1558 g / mL

ii) If the total mass of the prepared solution is 100 g, as previously calculated, then to know how many mL those grams equals, the density given is used as in the previous section:

100 g solution * (1 mL / 1.2185 g) = 82.07 mL solution

iii) In general, the wt% value of H2SO4 is calculated as follows:

wt% = (g pure H2SO4 / g Solution) * 100%

Knowing that the final value to be reached is 35%, and that this is based on 100 g of a solution containing 30 g of pure H2SO4 (that is, 30 wt% H2SO4) to which it is due to add an amount x in grams of a solution at 40 wt% H2SO4, so to solve this section you can consider the following equation:

35 wt% = ((30 g H2SO4 + (x * 40%)) / 100 g solution + x) * 100%

Where x would be the necessary grams of the solution at 40 wt% H2SO4. In the numerator, x is multiplied by 40% (0.4), considering that of the entire solution only 40% is pure H2SO4.

When clearing x, a value of 100 g is then obtained. Using the density of the solution at 40 wt% to convert them to volume, these would be:

100 g 40 wt% solution * (1 mL / 1,2991 g) = 76.98 mL of 40 wt% solution

the answer is na (sodium)