Respuesta :
Answer:
(a) 110 g/mol
(b) Ka = 2.14x10⁻⁵
Explanation:
(a)
The moles of OH⁻ at the equivalence point are equal to the moles of H⁺. Because the acid is monoprotic, the moles of H⁺ are equal to the moles of acid.
- moles OH⁻ ⇒ 0.0633 M * 18.4 mL = 1.16 mmol OH⁻ = 1.16 mmol H⁺ = 1.16 mmol Acid
Now that we have the moles of acid and its mass, we can calculate its molar mass:
- 127.6 mg / 1.16 mmol = 110 mg/mmol = 110 g/mol
(b)
We can use the Henderson-Hasselbach equation to solve this part:
- pH = pKa + log [A⁻]/[HA]
So we calculate the molar concentration of the basic (deprotonated, A⁻) and the acid (protonated, HA) forms:
The final volume is (25.0 mL + 10.0 mL) = 35.0 mL
- The amount of acid that turned into the deprotonated form is equal to the moles of OH⁻ added:
[A⁻] ⇒ 10.0mL * 0.0633 M / 35 mL = 0.018 M
- The amount of HA left unreacted is calculated from the original amount dissolved:
[HA] ⇒ (1.16 mmol acid - 10.0mL * 0.0633 M) / 35 mL = 0.015 M
Finally we calculate pKa and from it, Ka:
- pH = pKa + log [A⁻]/[HA]
- 5.87 = pKa + log (0.018 / 0.015)
- 5.87 = pKa + 1.20
- pKa = 4.67
Ka = [tex]10^{-4.67}[/tex] = 2.14 x 10⁻⁵
The molar mass of the unknown acid sample in the titration reaction is 110 g/mol. The acid dissociation constant is [tex]\rm 2.14\;\times\;10^{-5}[/tex].
What is titration?
- The titration is the neutralization reaction of acid and base to form a salt. The monoprotic aid has one hydrogen ion for each formula unit. Thus, the moles of a monoprotic acid are equivalent to the moles of hydrogen ions.
In the neutralization reaction, the acid is neutralized by the base. The moles of acid, are equivalent to the moles of base.
The given moles of the base are:
[tex]\rm Molarity=\dfrac{moles}{volume(L)} \\\\0.0633\;M=\dfrac{moles}{0.0184}\\\\ Moles=0.00116 moles\\Moles=1.16 mmoles[/tex]
The moles of the base in the neutralization are 1.16 mmoles. Thus, the moles of acid in the neutralization reaction are 1.16 mmoles.
The mass of the acid sample is 0.1276 grams. The molar mass of the acid can be given as:
[tex]\rm Molar\;mass=\dfrac{mass}{moles}\\\\ Molar\;mass=\dfrac{0.1276\;g}{0.00116\;mol}\\\\ Molar\;mass=110\;g/mol[/tex]
The molar mass of the unknown acid sample is 110 g/mol.
- The addition of 10 mL of base results in the final volume of the solution as:
[tex]\rm Final\;volume=25\;mL+10\;mL\\Final\;volume=35\;mL[/tex]
The pH of the solution determined is 5.87.
The molarity of the acid sample ([tex]\rm [A^-][/tex]) at 35 mL is given as:
[tex]\rm [A^-]=\dfrac{10\;mL\;\times\;0.0633\;M}{35\;mL}\\ A^-=0.0018\;M[/tex]
The concentration of the unreacted acid sample ([tex]\rm HA^-[/tex]) in the reaction can be given with the moles left as:
[tex]\rm HA^-=\dfrac{0.00116\;mol-10\;mL\;\times\;0.0633\;M}{35\;mL}\\ HA^-=0.015\;M[/tex]
The acid dissociation constant ([tex]\rm K_a[/tex]) for the reaction can be given as:
[tex]\rm pH=pK_a+log\;\dfrac{A^-}{HA^-}\\ 5.87=pK_a+log\dfrac{0.0018}{0.015} \\pK_a=4.67[/tex]
The acid constant can be derived as:
[tex]\rm K_a=10^{-pK_a}\\K_a=10^{-4.67}\\K_a=2.14\;\times\;10^{-5}[/tex]
The acid dissociation constant for the reaction is [tex]\rm 2.14\;\times\;10^{-5}[/tex].
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