Ask Question
31 January, 03:01

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?

0
Answers (1)
  1. 31 January, 04:25
    0
    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
Know the Answer?