Question 11 lon concentration refers to the molar concentration of an ion in sol

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Question 11 lon concentration refers to the molar concentration of an ion in solution. It may be identical to, or greater or less than, the molar concentration of the compound containing the ion that was used to make the solution. For soluble salts, the m olarity of a particular ion is equal to the m of that compound times the subscript for that ion. For example, 1 M of Alcia is 1 M in A13+ and 3 M in cl 1 M of (NHA) SOA is 2 M in NH and 1 M in 2- SO Submit Hints My Answers Give Up Review Part Part B lf CaCl2 is dissolved in water, what can be said about the concentration of the Ca2+ on? O It has the same concentration as the Cl ion. O Its concentration is half that of the Cl ion. O Its concentration is twice that of the Cl ion. O Its concentration is one-third that of the Cl ion. Submit Hints My Answers Give Up Review Part Part C A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams ofNa3PO4 will be needed to produce 550 mL of a solution that has a concentration of ons of 1.50 M? Express your answer to three significant figures and include the appropriate units. mass of Na3 PO4 3 Value Units Submit Hints My Answers Give Up Review Part

Explanation / Answer

Part B

Answer : The concentration of Ca++ is half that of the Cl-

Explanation:

CaCl2               =            Ca++    +    2 Cl-

1mol                             1 mol                 2 mol

1 mol CaCl2 ionizes to produce 1mol Ca++ and 2 mol Cl-

We find that,

the concentration of Cl- is twice that of the Ca++  

OR

the concentration of Ca++   is half that of the Cl-

PART C:

Na3PO4    =     3 Na +    +    PO43-

1 mol              3 mol   

0.5 mol          1.5 mol

Step 1: CALCULATION OF NUMBER OF MOLES OF Na3PO4

The number of moles of Na3PO4 needed to prepare 1000 mL of a solution which is 1.5M in Na + = 0. 5 mol

Therefore the number of moles of Na3PO4 needed to prepare 550 mL of a solution which is 1.5M in Na + = 0. 5 mol x 550 mL / 1000 mL = 0.275 mol

Step 2: CALCULATION OF NUMBER OF GRAMS OF Na3PO4

Molecular mass of Na3PO4 = 3 x 23 + 31 + 4 x 16 = 164 g/mol

1 mol Na3PO4 = 164 g of Na3PO4

Therefore, 0.275 mol Na3PO4 = (164 x 0.275) g of Na3PO4 = 45.1 g of Na3PO4

Answer :The number of grams of Na3PO4 that will be needed to produce 550 mL of a solution that has a concentration of Na + of 1.50 M is 45.1 g

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