When looking at the equilibrium between silver bromide and its aqueous ions, what could be added to solution to promote precipitation of silver bromide?

NaBr

Explanation:

When AgBr is dissolved in water, the following equilibrium is set up in solution;

AgBr(s)⇄Ag^+(aq) + Br^- (aq)

If we dissolve NaBr in the water, a common ion (Br^-) is now introduced into the system. This increases the concentration of Br^- and favours the reverse reaction hence more AgBr is precipitated. This is known as common ion effect.

a. pH is 2,88

b. pH is 8,87

Explanation:

a. Acetic acid (HC₂H₃O₂) dissociates in water thus:

HC₂H₃O₂ ⇄ C₂H₃O₂⁻ + H⁺; Ka = 1,8x10⁻⁵

ka =  [C₂H₃O₂⁻] [H⁺] / [HC₂H₃O₂] (1)

When you have a solution of 0,10M HC₂H₃O₂, the concentrations in equilibrium are:

[HC₂H₃O₂] = 0,10M - x

[C₂H₃O₂⁻] = x

[H⁺] = x

Replacing in (1)

1,8x10⁻⁵ = x² / 0,10 - x

1,8x10⁻⁶ - 1,8x10⁻⁵x = x²

x²+ 1,8x10⁻⁵x - 1,8x10⁻⁶ = 0

Solving for x:

x = - 0,00135067 -No physical sense, there are not negative concentrations-

x = 0,00133267  -Real answer-

As x = [H⁺] = 0,00133267

pH = -log [H⁺]

Thus, pH is 2,88

b. Sodium acetate (NaC₂H₃O₂) dissociates in water thus:

NaC₂H₃O₂ + H₂O ⇄ HC₂H₃O₂ + OH⁻ + Na⁺; Kb = 5,6x10⁻¹⁰

ka =  [HC₂H₃O₂] [OH⁻] / [NaC₂H₃O₂] (1)

When you have a solution of 0,10M NaC₂H₃O₂, the concentrations in equilibrium are:

[NaC₂H₃O₂] = 0,10M - x

[HC₂H₃O₂] = x

[OH⁻] = x

Replacing in (1)

5,6x10⁻¹⁰ = x² / 0,10 - x

5,6x10⁻¹¹ - 5,6x10⁻¹⁰x = x²

x²+ 5,6x10⁻¹⁰x - 5,6x10⁻¹¹ = 0

Solving for x:

x = -7,48359×10⁻⁶ -No physical sense, there are not negative concentrations-

x = 7,48303×10⁻⁶  -Real answer-

As x = [OH⁻] = 7,48303×10⁻⁶

pOH = -log [OH⁻]

pOH is 5,13. As 14 = pH + pOH

pH = 8,87

I hope it helps!