Question

What is the molarity of each ion in a solution prepared by dissolving 0.520 g of na2so4, 1.186 g of na3po4, and 0.223 g of li2so4 in water and diluting to a volume of 100.00 ml?

Answer 1

first calculate the number of moles of each ion present
for Na2So4 is
0.52/142=0.0037moles
Na+ =0.0037 x2 =0.0074moles
SO4^-2 =0.0027moes

for Na3PO4= 1.186/164=0.0072 moles
Na+ =0.0072 x3=0.0216moles
po4 ions = 0.0072 moles

for LiSo4 =0.223/110=0.002moles
Li ions=0.002 x2 =0.004mole
SO4 ions =0.002moles
total moles for
Na ions =0.0216 +0.0074 =0.029moles
forSO4 ions =0.0037 +0.002=0.0057moles
molarity is therefore
Na ion=0.029/0.1=0.29M
SO4 ions=0.0057/0.1=0.057M
Li ions =0.004/0.1=0.04M
PO4 ions=0.0072/0.1=0.072M