Suppose you prepare a solution that is 30wt% H2SO4 in water and intend to confir

Question

Suppose you prepare a solution that is 30wt% H2SO4 in water and intend to confirm the concentration by comparing the specific gravity of the solution to a value obtained from the literature. You carefully add 30g H2SO4 to 70.0g of water and allow the resulting mixture to equilibrate at 20 degrees celsius

a) from specific gravities in table B.1, estimate the volumes of water and sulfuric acid that have been blended and then estimate the density of the 30.0wt% sulfuric acid solution, assuming volumes are additive.

b) The value for 30.0wt% sulfuric acid at 20.0 degrees celsius is given 1.2185. Estimate the volume of the 30.0wt% solution prepared as descrbied by above process.

c) what volume of 40.0wt% sulfuric acid at 25 degrees celsius (SG=1.2991) must be added to 100.0g of 30.0wt% solution to produce a solution that is 35.0wt% sulfuric acid?

Explanation / Answer

(a) Given the mass of H2SO4 added = 30 g

Specific gravity of H2SO4 at 20 DegC = 1.839 g/mL

(Note: Use specific gravity value given in table -B1 to get even more accurate answer. If the value is same then no issue)

Hence volume of H2SO4, V1 = mass / specific gravity = 30 g / 1.839 g/mL = 16.31 mL (answer)

Given the mass of H2O added = 70.0 g

Specific gravity of H2O at 20 DegC = 0.998 g/mL

Hence volume of H2O, V2 = mass / specific gravity = 70.0 g / 0.998 g/mL = 70.14 mL (answer)

Now total mass of 30 wt% H2SO4 solution = 30.0 +70.0 = 100.0 g

Total volume of 30 wt% H2SO4 = V1+V2 = 16.31 mL + 70.14 mL = 86.45 mL

Hence density of 30 wt% H2SO4 solution = mass / volume = 100.0 g / 86.45 mL = 1.1567 g/mL (answer)

(c) Let the mass of 40% H2SO4 added = m g

Now total mass of H2SO4 after addition = 0.4 xm + 30 g

Now total mass of H2O after addition = 0.6 xm + 70 g

Hence total weight of 35.0wt% H2SO4 formed after mixing = 0.4 xm + 30 g + 0.6 xm + 70 g = (100 + m) g

Hence weight of H2SO4 in 35.0wt% H2SO4 = 0.35 x(100+m) g

Also 0.35 x(100+m) g = 0.4 xm + 30 g

=> 35 + 0.35m = 0.4m + 30

=> m = 5 / 0.05 = 100 g

Hence mass of 40.0wt% sulfuric acid at 25 degrees celsius (SG=1.2991) added = 100 g

Given SG = 1.2991

Hence volume of 40.0wt% sulfuric acid at 25 degrees celsius (SG=1.2991) added

= mass / SG = 100.0 g / 1.2991 =77.0 mL (answer)

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