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?

Question

Chemistry

Angelica Thomas

27 March, 02:42

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?

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Short · 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