Answer:
Products are favored.
Explanation:
The acid-base reaction of CH₃COOH (acid) with NH₃ (base) produce:
CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = ?
It is possible to know Kr of the reaction by the sum of acidic dissociations of the half-reactions. That is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Ka = 1.8x10⁻⁵
NH₃ + H⁺ ⇄ NH₄⁺ 1/Ka = 1/ 5.6x10⁻¹⁰ = 1.8x10⁹
___________________________________
CH₃COOH + NH₃ ⇄ CH₃COO⁻ + NH₄⁺ Kr = 1.8x10⁻⁵×1.8x10⁹ = <em>3.2x10⁴</em>
<em> </em>
As Kr is defined as:
Kr = [CH₃COO⁻] [NH₄⁺] / [CH₃COOH] [NH₃]
And Kr is > 1
[CH₃COO⁻] [NH₄⁺] > [CH₃COOH] [NH₃],
showing <em>products are favored</em>.
To solve this problem,
we can use the Henderson-Hasselbalch Equation which relates the pH to the measure
of acidity pKa. The equation is given as:<span>
<span>pH = pKa + log ([base]/[acid]) ---> 1</span></span>
Where,
[base] = concentration
of C2H3O2
in molarity or moles
<span>[acid] = concentration of HC2H3O2 in molarity or moles</span>
For the sake of easy calculation, let us assume that:
[base] = 1
[acid] = x
<span>
Therefore using equation 1,
4.24 = 4.74 + log (1 / x)
<span>log (1 / x) = - 0.5
1 / x = 0.6065 </span></span>
x =
1.65<span>
The required ratio of C2H3O2 /HC2H3O2 <span>
is 1:1.65 or 3:5. </span></span>
Usually in this context you would be referring to the boiling and freezing point of a NaCl <em>solution</em> (saltwater) compared to pure H_{2}O. Sematics would be different for NaCl compound itself, you would say melting and boiling point for a solid substance- and the temperatures would be very, very radical (high).
The boiling point of pure water is 100 degrees C (212 F), and the freezing/melting point is below 0 degrees C (32 F). For a salt water solution, the boiling point is raised and the melting point is lowered. This means that water will stay liquid for an increased range of temperature. Depending on the amount of NaCl solute in the water, the boiling and melting points may change a few degrees.
With the info given i would have to say their is no kinetic energy, it's all potential energy.
