Arsenic acid has the formula H^3AsO^4. It is a triprotic acid. Abottle is marked 0.422 N arsenic acid. What is the Molarity of this solution?
M = N/3
To find the molarity of a solution, we need to know the amount of solute (in moles) and the volume of the solution (in liters).
In this case, we are given that the solution is marked as 0.422 N arsenic acid. N stands for normality, which is a measure of chemical concentration. To convert N to molarity, we need to know the equivalent weight of the acid.
The molecular weight of arsenic acid (H3AsO4) is:
(3 x 1 atomic mass of hydrogen) + (1 x 1 atomic mass of arsenic) + (4 x 1 atomic mass of oxygen) = 3 + 74.92 + 64 = 141.92 g/mol
Since arsenic acid is a triprotic acid, it can donate three moles of hydrogen ions (H+) per mole of acid. Therefore, the equivalent weight of arsenic acid is one-third of its molecular weight:
Equivalent weight = Molecular weight / Number of equivalents per mole
Equivalent weight = 141.92 g/mol ÷ 3 equivalents/mol = 47.31 g/equivalent
Now, let's calculate the molarity of the solution:
Molarity (M) = Normality (N) × Equivalent weight (g/equivalent) / 1000 (to convert g to kg)
Molarity = 0.422 N × 47.31 g/equivalent ÷ 1000 = 0.01998 M
Therefore, the molarity of the solution is approximately 0.01998 M.