50.00 mL of a 0.10 M NaCL is added to 75 mL of a 0.25 M AgNO3 solution. Write th

Question

50.00 mL of a 0.10 M NaCL is added to 75 mL of a 0.25 M AgNO3 solution. Write the balanced chemical equation for this reactions.

(a). Write the balanced chemical equation for this reaction
(b). Write the total ionic and net ionic equations
(c). What are the spectators
(d). What precipitate is formed?
(e). What is the concentration of the sodium ion in the resulting solution?
(f). What is the concentration of the nitrate ion in the resulting solution?
(g). What is the approximate concentration of the chloride ion the resulting solution?
(h). What is the approximate concentration of the silver ion the resulting solution?

Explanation / Answer

a) NaCl (aq) + AgNO3 (aq) -----------------> AgCl(s) + NaNO3(aq)

b) Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) ------------->  Na+(aq) + NO3-(aq) + AgCl(S)

net ionic equation Cl-(aq) + Ag+(aq) -------------> AgCl(S)

C) the spectator ions are Na+ and NO3-

d)AgCl is the precipitate formed

e)

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

50x0.1 =5 75x 0.25 =18.75 0 0 initial concentration

All the sodium remains in solution .

So [Na+] = mmoles / total volume

= 5/125 =0.04M

f) All the nitrate also remains in solution as it is a spectator ion.

[NO3-] = 18.75/125

=0.15 M

g) As the [Ag+] >> [Cl-] all the chloride ion is precipitated thus the approximate concentration of [Cl-] = 0

h)

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

50x0.1 =5 75x 0.25 =18.75 0 0 initial concentration

0 13.75 5 - after reaction

thus [Ag+] remaining = 13.75 /125

=0.11 M

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.