What is the concentration (M) of sodium ion in a solution made by mixing 100. mL
ID: 1000852 • Letter: W
Question
What is the concentration (M) of sodium ion in a solution made by mixing 100. mL of 0.100 M sodium sulfate, 100. mL of 0.50 sodium chloride, and 200. mL of 0.0250 M sodium phosphate? (Assume additive volumes) What is the concentration (M) of sodium ion in a solution made by mixing 100. mL of 0.100 M sodium sulfate, 100. mL of 0.50 sodium chloride, and 200. mL of 0.0250 M sodium phosphate? (Assume additive volumes) What is the concentration (M) of sodium ion in a solution made by mixing 100. mL of 0.100 M sodium sulfate, 100. mL of 0.50 sodium chloride, and 200. mL of 0.0250 M sodium phosphate? (Assume additive volumes)Explanation / Answer
100. mL of 0.100 M sodium sulfate
no ofmoles of sodium sulfate = molarity * volume in L
= 0.1*0.1 = 0.01 moles of Na2So4
no of moles of Na+ = 2* 0.01 = 0.02 moles of Na+
100. mL of 0.50 sodium chloride
no of moles of NaCl = molarity * volume in L
= 0.5*0.1 = 0.05 moles of NaCl
no of moles of Na+ = 1*0.05 = 0.05 moles
200. mL of 0.0250 M sodium phosphate
no of moles of Na3PO4 = 0.025*0.2 =0.005 moles of Na3PO4
no of moles of Na+ = 3*0.005 = 0.015 moles
Total no of moles of Na+ = 0.02 + 0.05 + 0.015 = 0.085 moles
Total volume = 100+ 100+200 = 400ml = 0.4 L
molarity of Na+ = no of moles/volume in L
= 0.085/0.4 = 0.2125M
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