calculate the molarity of sodium ion in a solution made by mixing 3.4mL of .386 M soldium chloride with 475mL?

calculate the molarity of sodium ion in a solution made by mixing 3.4mL of .386 M sodium chloride with 475mL of .0712 M sodium sulfate. assume volume is additive

2 Answers

  • Determine the total number of moles of Na+ you add

    Then determine the conc of this many moles in the total volume

    moles = molarity x Litres

    moles NaCl = 0.386 M x 0.0034 L

    moles NaCl = 0.0013124 moles

    Each NaCl has 1 Na+

    Therefore moles Na+ from NaCl = 0.0013124 moles

    moles Na2SO4 = 0.0712 M x 0.475 L

    moles Na2SO4 = 0.03382 moles

    Each Na2So4 has 2 Na+

    Therefore moles Na+ from Na2SO4 = 2 x 0.03382 = 0.06764 moles Na+

    Total moles Na+ = 0.06764 + 0.0013124 = 0.0689524 moles Na+

    Total volume = 475 ml + 3.4 ml = 478.4 ml = 0.4784 L

    molarity Na+ = moles / L

    = 0.0689524 mol / 0.4784 L

    = 0.14 M (2 sig figs)

  • Your very final concentration of Na+ is a million.496 x 10-2 molar. that’s the way you hit upon it out: Your first answer is 0.280 mole Na+/one hundred mL x 3.68 mL = a million.03 x 10-3 mole Na+ Your 2nd answer is 6.51 x 10-3 mol Na+/one thousand mL x 500 mL X 2 Na+/Na2SO4 formulation unit = 6.51 x 10-3 moles Na+. consisting of them on an identical time provides you 7.fifty 4 x 10-3 moles Na+/0.50368 L, which equals a million.496 x 10-2 moles Na+ in step with liter or a million.496 x 10-2 molar in Na+.

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