How to calculate the following amounts of substance needed to prepare a solution?

Question

I'm having difficulty with this question:

Two solutions, 0.10L of 0.12M NaCl, and 0.23L of 0.18M MgCl2, are mixed together. What volume (in mL) of a 0.20M solution of AgNO3 is needed to precipitate all the Cl- ion as AgCl?

I'm not sure how I need to go about doing this.

Answer 1

Take them one at a time and add up the moles of chloride

solution 1)

0.1L of .12M NaCl

0.1L * 0.12 mole/L = 0.012 mole Cl-

solution 2)

0.23l of 0.18M MgCl2

MgCl2 --> Mg+2 + 2Cl-

0.23L * (0.18mole/L) (2Cl-/molecule) = 0.0828 mole Cl-

total Cl- = 0.012 + 0.0828 = 0.0948 moles

You need to ppt this many moles with 0.2M AgNO3

xL * 0.2 moleAgNO3/L = 0.0948 mole

xL = 0.474 L

You will need 474 ml of 0.2M AgNO3

Answer 2

You solve it for the two solutions separately, then add up the amounts. Let the NaCl solution be called NaS. Let the MgCl2 solution be called MgS. Let the AgNO3 solution be called AgS.

0.10LNaS x 0.12molNaCl/1LNaS x 1molCl-/1molNaCl = 0.012mol Cl-

0.23LMgS x 0.18molMgCl2/1LMgS x 2molCl-/1molMgCl2 = 0.0828molCl-

0.012 + 0.0828 = 0.0948 moles Cl- total

0.0948molCl- x 1molAgNO3/1molCl- x 1LAgS/0.20molAgNO3 = 0.47 L AgS required (to two significant figures)

Answer 3

0.1 liters of 0.12 M NaCl = 0.012 moles NaCl = 0.012 equivalents of NaCl
.23 liters of 0.18 M MgCl2 = 0.0414 moles MgCl2 = 0.0828 equivalents

Total equivalents = 0.012 + 0.0828 = 0.0948 = moles of chloride ion

It takes 1 silver ion to precipitate 1 chloride ion.

AgNo3 equivalents needed = 0.0948 = moles of AgNo3

Solution of 0.2 M AgNO3 needed = 0.0948/0.2 = 0.474 liters

Answer 4

whats up in simple terms kno this one factor and all ur doubts of molarity would be cleared. assume we are speaking approximately molarity of A then in a answer of quantity V then Molarity=Moles of A/V(in litres) and for pK factor pK= -log (molarity of ok) comparable for Cl-

Answer 5

((0.1 x .12) + (2 x 0.23 x 0.18))/0.20 = 0.474