The normality of Hv2SOv4 is given as 0.20 N for a reaction in which the Hv2SOv4 furnishes two hydrogen ions. What is the molarity of the Hv2SOv4?
A. 0.40 M
B. 0.30 M
C. 0.20 M
D. 0.10 M
Look at it this way.
98 g H2SO4 = 1M = 2N; therefore,
M x #Hatoms = N
Rearrange, since you want M to]
M = N/#H atoms = 0.2/2 = ?? N
To determine the molarity of Hv2SOv4, we need to understand the relationship between normality and molarity.
Normality (N) is a measure of concentration that takes into account the number of equivalents of a substance involved in a reaction. Equivalents are a measure of the reactive capacity of a compound. For acids and bases, normality is equal to the molarity multiplied by the number of hydrogen ions (H+) or hydroxide ions (OH-) produced or consumed in the reaction.
In this case, the normality of Hv2SOv4 is given as 0.20 N, and it furnishes two hydrogen ions (H+). Therefore, we can set up the following equation to relate normality and molarity:
Normality (N) = Molarity (M) × Number of hydrogen ions (H+)
0.20 N = M × 2 H
Now, we can solve for the molarity (M):
M = 0.20 N / 2 H
M = 0.10 N/H
Since the number of hydrogen ions (H+) is 2, we divide 0.10 N by 2:
M = 0.10 N / 2
M = 0.05 N
Therefore, the molarity of Hv2SOv4 is 0.05 M.
The correct answer choice is not provided in the options given.