calcuate the normality of 0.2M solution of h3p04 ?
The Normality of H3PO4 (f.wt. = 98 g/mol) depends upon what type of reaction it is associated with. If it is in a metathesis reaction, the equivalent weight is the mass of reagent delivering 1 mole of + charge.
=> H3PO4 => 3H^+ + PO4^- => Eqv Wt = (98g/mol)/(3mole +) = 32.7g/eqv wt.
Given 0.20M H3PO4 => (0.20mol/L)(98g/mol)=1.96g/L => N(0.20M H3PO4)=(1.96g/L)/(32.7g/eqv wt)=0.60N H3PO4
If the reagent is in an oxidation-reduction reaction, however, the equivalent weight is the weight of reagent that can deliver or gain 1 mole of electrons in the redox process.
Here's an example ...
3P4 + 5HNO3 +38H2O => 12H3PO4 + 5NO + 45H^+
P4 => 4H3PO4 + 4(5e^-)
(1/4)P4 => H3PO4 + 5e^-
1 eqv wt in this redox process = [98g/5]=19.6g. => Normality of a 0.20M H3PO4 solution = (0.20mol/L)(98g/mol)/(19.6g/eqv wt) = 0.10eqv wts/Liter = 0.10N H3PO4
To calculate the normality of a solution, you need to know the number of acidic or basic equivalents present in the solute.
For phosphoric acid (H3PO4), it has three acidic hydrogens, so it can donate three acidic equivalents in a reaction.
To determine the normality, you can use the formula:
Normality (N) = (Molarity (M)) * (Number of Acidic Equivalents)
In this case, the molarity of the solution is given as 0.2M, and the number of acidic equivalents is 3.
Plugging in these values:
Normality (N) = (0.2M) * (3) = 0.6N
Therefore, the normality of a 0.2M solution of H3PO4 is 0.6N.
To calculate the normality of a solution, you need to know the number of equivalents of the solute present in the solution. In this case, the solute is H3PO4 (phosphoric acid).
Phosphoric acid (H3PO4) is a triprotic acid, meaning it can donate three protons (H+) per molecule. Thus, to find the number of equivalents of H3PO4, multiply the concentration (molarity) of the acid by the number of protons it can donate.
In this case, the concentration of the solution is 0.2 M, which means there is 0.2 moles of H3PO4 per liter of solution.
Since H3PO4 is a triprotic acid, it can donate three protons per molecule. Therefore, the number of equivalents of H3PO4 is 0.2 M x 3 equivalents/mole = 0.6 equivalents.
So, the normality of the 0.2 M solution of H3PO4 is 0.6 N.