Question

What is the result of mixing 25.0 mL of 0.0010 M NaCl solution with 75.0 mL of 0.0050 M AgNO₃ solution? Assume volumes are additive.

Answer 1

Molarity of new solution is 0:36M__ Ag Noslat nachlag) Agilis) + ali + NaNO3- moles of AgNO3= Molarity & volume (4) 0.0050 x 75 x10-3 = 3,75 x10 Smoles I moles of Nace = Molarity x volune (2) = 0.0010X 25 X10-3: 205 x10~ Smolee from we with the stoichiometry of can see that I mol 1 mol of Mace to reaction equation of AgNO3 reacts form I mol Agel. So, . . Here moles Nace of is All the formed limiting reagent. = 2.5x10 moles

^ Moles of Moles of All formed= 2.5x10-s moles. AgNO3 left = 3.75*10" - 2.5x105 - 3.5810-4 moles. Moles of Nano3 formed = 2.5x10-5 moles. moles of nace left = 0.
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