Calculate the concentration of the ions in 2.13mol/L solution of Li3PO4(aq). How many of each ion would be present in 2500mL of this solution?
ANSWERS:
[Li+1]= 6.39mol/L
[PO4-3]= 2.13mol/L
# Li+1=9.62x1024
#PO4-3= 3.21x1024
If you have a 1.2L of a 2.13mol/L solution of Li3PO4(aq) and you remove 25mL of this solution and add it to 150mL of water, what is the new concentration of ions in the diluted solution?
ANSWERS:
[Li3PO4]= 0.30mol/L
[Li+1]= 0.91mol/L
[PO4-3]= 0.30mol/L
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FLOW CHART REVIEW
# of moles
of Atoms/ions
ly
tip by the number
ul
m
# of Atoms
# of molecules
of atoms or
ions per
molecule
div
ide
6.02x1023
div
ide
multiply
by the number
of atoms or
ions per
molecule
# of Atoms/ions
Moles
How many moles of Hydrogen atoms/ions are
in the following samples?
152mL of H2O
Answer: 152mL x 1g/mL divided by 18.02g/mol
= 8.44mol H2O
# mole H = 8.44mol x2
# mole O = 8.44mol x1
152g of H3PO4
Answer: 152g divided by 98.00g/mol
= 1.55 mole H3PO4
+1
# mole H = 1.55mol x3
# mole PO4-3 = 1.55mol x1
How many Hydrogen atoms/ions are in the
following samples?
152mL of H2O
Answer: 152mL x 1g/mL divided by 18.02g/mol
= 8.44mol H2O
# H = 8.44mol x 6.02x1023 x2 =1.02x1025
# O = 8.44mol x 6.02x1023 x1 =5.08x1024
152g of H3PO4
Answer: 152g divided by 98.00g/mol
= 1.55 mole H3PO4
# H+1 = 1.55mol x 6.02x1023 x3 =2.80x1024
# PO4-3 = 1.55mol x 6.02x1023 x1 =9.33x1023
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