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Predict whether or not a precipitate will form upon mixing 175.0 ml of a 0.0055 mkcl solution with 145.0 ml of a 0.0015 m agno3 solution. identify the precipitate, if any. express your answer as a chemical formula. enter noreaction if no precipitate is formed.



Answer :

znk

Answer:

AgCl

Step-by-step explanation:

The equation for a possible reaction is

KCl(aq) + AgNO₃(aq) ⟶ KNO₃ + AgCl

The possible precipitate is silver chloride.

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Moles KCl = 0.1750 × 0.0055/1

Moles KCl = 9.62 × 10⁻⁴ mol

Moles AgNO₃ = 0.1450 × 0.0015

Moles AgNO₃ = 2.18 × 10⁻⁴  mol

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Total volume = 175.0 + 145.0

Total volume = 320.0 mL

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[KCl] = 9.62 × 10⁻⁴/0.3200

[KCl] = 3.01 × 10⁻³ mol·L⁻¹

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[AgNO₃] = 2.18 × 10⁻⁴/0.3200

[AgNO₃] = 6.80 × 10⁻⁴ mol·L⁻¹

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The equation for the equilibrium is

                AgCl ⇌        Ag⁺       +      Cl⁻

I/mol·L⁻¹:                  6.80 × 10⁻⁴   3.01 × 10⁻³

Q_sp = [Ag⁺][Cl⁻]

Q_sp = 6.80 × 10⁻⁴ × 3.01 × 10⁻³

Q_sp = 2.04 × 10⁻⁶

K_sp = 1.8 × 10⁻¹⁰

Q_sp > K_sp, so a precipitate will form.

Cricetus

"[tex]Qsp>Ksp[/tex]" precipitate will form. A further explanation is below.

According to the question,

  • [tex]AgNO_3(aq) +KCl(aq) \rightarrow AgCl(s)+KNO_3(aq)[/tex]

We can say the,

AgCl is a precipitate.

Ksp of AgCl,

  • [tex]1.8\times 10^{-10}[/tex]

then,

→ [tex]Qsp = [Ag^+][Cl^-][/tex]

By substituting the values, we get

→         [tex]= 0.0015\times 0.0055[/tex]

→         [tex]= 8.25\times 10^{-6}[/tex]

Thus the above answer is appropriate.

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