A gas container has a volume of 446.9 with a temp of 14c. When the volume is decreased to 238.7l the new temp is what
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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>