Question

What is the Normality of a solution made by diluting 50.00 ml of H2SO4 with a specific gravity of 1.080 containing 11.60% H2SO4 by weight to one liter with water?

Answer 1

Answer on the question #65902, Chemistry / Other

Question:

What is the Normality of a solution made by diluting $50.00\mathrm{ml}$ of H2SO4 with a specific gravity of 1.080 containing $11.60\%$ H2SO4 by weight to one liter with water?

Solution:

Normality, or equivalent concentration, is molar concentration, divided by the factor of equivalency:

$$N = \frac {c _ {i}}{f _ {e q}}$$

As one molecule of sulfuric acid produces two hydronium ions in the solution, factor of equivalency is 0.5.

Molar concentration is:

$$c _ {a c i d} = \frac {n (a c i d)}{V (s o l u t i o n)} = \frac {m _ {a c i d}}{M _ {a c i d} V _ {s o l .}}$$

Volume of the solution $V_{sol}$ is 1L, molar mass is 98,0785 g/mol. The mass of acid added is (taking the density of water as 1 g/ml, the density of stock solution is 1.080 g/ml):

$$m _ {a c i d} = m _ {s t o c k s o l.} \omega_ {s t o c k s o l.} = V _ {s t o c k s o l.} d _ {s t o c k s o l.} \omega_ {s t o c k s o l.}$$ $$m _ {a c i d} = 5 0. 0 (m l) \cdot 1. 0 8 0 (g m l ^ {- 1}) \cdot 0. 1 1 6 0 = 6. 2 6 4 g$$ $$c _ {a c i d} = \frac {6 . 2 6 4 (g)}{9 8 . 0 7 8 5 (g m o l ^ {- 1}) \cdot 1 (L)} = 0. 0 6 3 8 7 m o l L ^ {- 1}$$ $$N = \frac {c}{f} = \frac {0 . 0 6 3 8 7 m o l L ^ {- 1}}{0 . 5} = 0. 1 2 7 7 e q L ^ {- 1}$$

Answer: 0.1277 eq/L

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