Would a precipitate form if a solution contained 8.0 x 10-4 m silver nitrate and 1.8 x 10-4 m potassium chromate? you will have
1 answer:
You might be interested in
The major species in solution when solid ammonium bromate is dissolved in water is shown below
The solution is as follows:
The reaction is written in the attached picture.
Mol NH₃: 0.10 mol/L * 100 mL * 1 L/1000 mL = 0.01 mol
Mol HNO₃: 0.10 mol/L * 150 mL * 1 L/1000mL = 0.015 mol
Mol NH₄NO₃ produced: 0.01 mol NH₄NO₃
Mol HNO₃ left = 0.015 - 0.01 = 0.005 mol
Hydrolyzing NH₄⁺ and applying ICE approach
NH₄⁺ --> H⁺ + NO₃⁻
I 0.01 0 0
C -x +x +x
E 0.01-x x x
Kh = Kw/Kb = [H⁺][NO₃⁻]/[NH₄⁺]
10⁻¹⁴/1.8×10⁻⁵ = [x][x]/[0.01-x]
Solving for x,
x = [H⁺] = 2.357×10⁻⁶ mol
The formula for pH is
pH = -log [H⁺]
Aside from 2.357×10⁻⁶ mol, let's add the H⁺ from the remaining HNO₃ which is 0.005.
Therefore,
pH = -log[2.357×10⁻⁶ mol + 0.005 mol]
<em>pH = 2.3</em>
The reaction is written in the attached picture.
Mol NH₃: 0.10 mol/L * 100 mL * 1 L/1000 mL = 0.01 mol
Mol HNO₃: 0.10 mol/L * 150 mL * 1 L/1000mL = 0.015 mol
Mol NH₄NO₃ produced: 0.01 mol NH₄NO₃
Mol HNO₃ left = 0.015 - 0.01 = 0.005 mol
Hydrolyzing NH₄⁺ and applying ICE approach
NH₄⁺ --> H⁺ + NO₃⁻
I 0.01 0 0
C -x +x +x
E 0.01-x x x
Kh = Kw/Kb = [H⁺][NO₃⁻]/[NH₄⁺]
10⁻¹⁴/1.8×10⁻⁵ = [x][x]/[0.01-x]
Solving for x,
x = [H⁺] = 2.357×10⁻⁶ mol
The formula for pH is
pH = -log [H⁺]
Aside from 2.357×10⁻⁶ mol, let's add the H⁺ from the remaining HNO₃ which is 0.005.
Therefore,
pH = -log[2.357×10⁻⁶ mol + 0.005 mol]
<em>pH = 2.3</em>