how to calculate the weight of solute req'd to prepare 1 liter of 1 normal of:
a.) LiOH
b.) phosphoric acid
my work:
formula= gew/L or w/ewxL
i get their mw then divide it to their H/OH so i can get EW then multiply it to 1.
a.) 24 g/mol/1
b.) 98g/mol/3
i don't know what's next to find to be able to follow the formula. how to find the weight!! please help me!
You're there.
Since 24/1 is the EW for LiOH, then 24 g LiOH in 1L solution will be 1 N. (also 1 M)
b. Same for H3PO4. Dissolve 98/3 g H3PO4 in a little water and make to the 1L mark with water to prepare a 1 N solution of H3PO4. Technically you need the reaction being used in H3PO4 since there COULD be 3 different equivalent weights depending upon the number of H ions involved but USUALLY in these questions the maximum is used. So 98/3 is the g required. (By the way that makes a 0.333 M solution)
what am i going to do next? please heelp.
To calculate the weight of solute required to prepare 1 liter of a 1 normal solution, you need to follow these steps:
Step 1: Determine the formula weight (FW) of the solute.
The formula weight is the sum of the atomic weights of all the atoms in the formula unit of the compound. You correctly calculated the formula weight (FW) for LiOH to be 24 g/mol and for phosphoric acid (H3PO4) to be 98 g/mol.
Step 2: Determine the equivalent weight (EW) of the solute.
The equivalent weight represents the weight of the solute that can react with or replace one mole of hydrogen ions (H+) or hydroxide ions (OH-) in the solution. To determine the equivalent weight, divide the formula weight (FW) by the number of H+ or OH- ions that are produced or consumed during the reaction.
a) For LiOH: The reaction of LiOH in water produces one hydroxide ion (OH-). Therefore, the equivalent weight of LiOH is equal to its formula weight (FW), which is 24 g/mol.
b) For phosphoric acid (H3PO4): The reaction involves the donation of 3 hydrogen ions (H+) from phosphoric acid, so the equivalent weight (EW) will be the formula weight (FW) divided by 3. Hence, the EW of phosphoric acid is 98 g/mol divided by 3, which is approximately 32.67 g/mol.
Step 3: Calculate the weight of solute required.
Using the formula weight (FW) and the equivalent weight (EW), we can now find the weight of solute required. So, using the general formula:
Weight = [Equivalent Weight (EW) x Normality (N)] x Volume (V)
For a 1 normal (1N) solution with a volume of 1 liter, the weight of solute required can be calculated as:
a) For LiOH:
Weight = [24 g/mol x 1 N] x 1 L = 24 grams of LiOH
b) For phosphoric acid:
Weight = [32.67 g/mol x 1 N] x 1 L = 32.67 grams of phosphoric acid
Therefore, to prepare a 1 normal solution of LiOH, you need 24 grams of LiOH, and to prepare a 1 normal solution of phosphoric acid, you need 32.67 grams of phosphoric acid.