If 8.5g of pure ammonium phosphate , (NH4)3PO4(s), is dissolved in distilled water to make 400mL of solution, what are the concentrations (in moles per litre) of ions in the solution.
How would I solve this type of stoichiometric problem. I know the steps
Step 1: make the balanced equation and state the givens and unknowns
Step 2: find the moles using c=n/v
Step 3: find the mole ratio
Step 4: find the mass
Please guide me! Thanks!!:) I appreciate your time!
What’s the answer
To solve this stoichiometric problem, you are on the right track with the steps you provided. Here is a step-by-step breakdown of how to solve it:
Step 1: Make the balanced equation and state the givens and unknowns.
The balanced equation for the dissociation of (NH4)3PO4(s) is as follows:
(NH4)3PO4(s) → 3 NH4+(aq) + PO4^3-(aq)
Given:
- Mass of ammonium phosphate = 8.5g
- Volume of solution = 400mL (which is equivalent to 0.4L)
Unknowns:
- Concentration of NH4+ ions (in moles per litre) in the solution
- Concentration of PO4^3- ions (in moles per litre) in the solution
Step 2: Find the moles using c=n/v.
To calculate the moles of (NH4)3PO4, divide the given mass by its molar mass:
Molar mass of (NH4)3PO4 = (3 x molar mass of NH4+) + molar mass of PO4^3-
= (3 x (1 x atomic mass of H + atomic mass of N)) + atomic mass of P + (4 x atomic mass of O)
Step 3: Find the mole ratio.
From the balanced equation, we can see that 1 mole of (NH4)3PO4 dissociates to give 3 moles of NH4+ ions and 1 mole of PO4^3- ions.
Step 4: Find the concentration (in moles per litre).
To find the concentration of NH4+ and PO4^3- ions in the solution, divide the moles of each ion by the volume of the solution (in litres).
NH4+ concentration = (moles of NH4+ ions) / (volume in litres)
PO4^3- concentration = (moles of PO4^3- ions) / (volume in litres)
Following these steps, you should be able to calculate the concentrations of NH4+ and PO4^3- ions in the solution.
To solve this stoichiometric problem, you're on the right track with the steps you mentioned. Let's break it down further with a step-by-step guide:
Step 1: Balanced Equation and Givens
The given substance is pure ammonium phosphate, (NH4)3PO4(s), which dissociates into ions in water: 3NH4+ and PO4-3. The balanced equation is:
(NH4)3PO4(s) → 3NH4+(aq) + PO4-3(aq)
Step 2: Finding Moles
First, calculate the number of moles of (NH4)3PO4 using the given mass and the molar mass of (NH4)3PO4. The molar mass of (NH4)3PO4 is:
(NH4)3PO4 = (3x NH4) + PO4
= (3x 18.03 g/mol) + 94.97 g/mol
= 55.09 g/mol
Now, using the formula c = n/V (concentration = moles/volume), you can calculate the concentration of (NH4)3PO4:
c = (8.5 g)/(0.4 L) = 21.25 g/L
Since we want the concentration in moles per liter (mol/L), convert grams to moles by dividing by the molar mass:
n = (8.5 g)/(55.09 g/mol) = 0.154 mol
Step 3: Mole Ratio
From the balanced equation, you can see that 1 mole of (NH4)3PO4 dissociates into 3 moles of NH4+ ions. Therefore, the concentration of NH4+ ions is 3 times the concentration of (NH4)3PO4:
Concentration of NH4+ = 3 x (0.154 mol/L) = 0.462 mol/L
Similarly, the concentration of PO4-3 ions is also 0.154 mol/L, as 1 mole of (NH4)3PO4 dissociates into 1 mole of PO4-3 ions.
Step 4: Mass
If you want to find the mass of each ion, you can multiply the moles by their respective molar masses. The molar mass of NH4+ is 18.03 g/mol, so the mass of NH4+ is:
Mass of NH4+ = (0.462 mol/L) x (18.03 g/mol) = 8.34 g/L
For PO4-3, the molar mass is 94.97 g/mol; thus, the mass of PO4-3 is:
Mass of PO4-3 = (0.154 mol/L) x (94.97 g/mol) = 14.62 g/L
So, the concentrations (in moles per liter) of ions in the solution are:
- NH4+: 0.462 mol/L
- PO4-3: 0.154 mol/L
And the corresponding mass of each ion is:
- NH4+: 8.34 g/L
- PO4-3: 14.62 g/L
Keep in mind that these calculations assume complete dissociation of (NH4)3PO4 in water.