Calculate the molarity of sodium ion in a solution made by mixing 5.89 mL of 0.130...
Calculate the molarity of sodium ion in a solution made by mixing 5.89 mL of 0.130 M sodium chloride with 475.00 mL of 7.27 × 10−2 M sodium sulfate. Assume volumes are additive.
Homework Answers
Moles of NaCl = molarity*volume in L
= 0.130M*0.00589L = 7.66*10-4mol
Moles of Na+ = 7.66*10-4 mol
Moles ofNa2SO4 = 7.27*10-2M*0.475L = 345.3*10-4mol
Moles of Na+ = 2*345.3*10-4 = 6.91*10-4mol
Total moles of Na+ = 7.66*10-4mol + 6.91*10-4mol
= 14.57*10-4 mol
Total Volume = 0.00589L + 0.475L = 48.1*10-2L
Molarity of Na+ ions = 14.57*10-4mol/48.1*10-2L
= 3.03 *10-3 M
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