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Question

Calculate the pCl of the mixture of 100.0 mL of 0,0500 M NaCl with 40.0 mL 0.100 M AgNO3 . Take account of the precipitate occurs when the solutions are mixed.

PLEASE WRITE THE SOLUTION WITH FULL STEPS

Answer 1

NaCl(aq) + AgNO₃(aq) ==> AgCl(s) + NaNO₃(aq)

moles NaCl present = 100.0 ml x 1 L/1000 ml x 0.0500 mol/L = 0.00500 moles NaCl

moles AgNO₃ present = 40.0 ml x 1 L/1000 ml x 0.100 mol/L = 0.00400 moles AgNO₃

moles NaCl left over after reaction = 0.00500 - 0.00400 = 0.00100 moles NaCl left over = 0.00100 moles Cl^-

Final volume = 100.0 ml + 40.0 ml = 140 ml = 0.140 L

[Cl-] = 0.00100 moles/0.140 L = 0.00714 M

pCl = -log Cl- = -log 0.00714 = 2.15