Answer:
pH = 5.76
Explanation:
We can solve this problem by using<em> Henderson-Hasselbach's equation</em>:
pH = pKa + log![\frac{[SodiumAcetate]}{[AceticAcid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSodiumAcetate%5D%7D%7B%5BAceticAcid%5D%7D)
We are already know all the required information, thus we<u> input the data given by the problem</u>:
pH = 4.76 + log(20/2)
And finally <u>calculate the pH</u>:
pH = 5.76
The pH of that acetic acid solution is 5.76.
I believe it is BaS I hope this helps!
Answer:
E) Two of the above statements are true.
Explanation:
The options are:
A) Before the solution is titrated with HCl it is pink and when the color changes from pink to colorless, the moles of H*(aq) equals the moles of OH"(aq) used in the hydrolysis of the neutralized aspirin. <em>TRUE. </em>Before the solution is titrated, there is an excess of NaOH (Basic solution, phenolphtalein is pink). Then, at equivalence point, after the addition of HCl, the pH is acidic and phenolphtalein is colorless.
B) Before the solution is titrated with HCl it is colorless and when the color changes from colorless to pink, the moles of H*(aq) equals the excess moles of OH(aq) added. <em>FALSE. </em>As was explained, before the titration, the solution is pink.
C) 25.0 mL of 0.100 M NaOH was added to the sample to hydrolyze the neutralized aspirin in the solution. The titration with HCl allows us to determine the moles of excess OH(aq) added. Once we determine the moles of excess OH(aq), we can determine moles of OH"(aq) used in the hydrolysis of the neutralized aspirin, which is equal to the moles of aspirin in the recrystallized aspirin. <em>TRUE. </em>Aspirin requires an excess of base (NaOH) for a complete dissolution (Hydrolysis). Then, we add H+ as HCl to know the excess moles of OH-. As we know the added moles of OH-, we can find the moles of OH that reacted = Moles of aspirin.
D) We can determine the moles of aspirin in the recrystallized aspirin by titrating with the 0.100 M NaOH to the neutralization point. The purpose of the hydrolysis of the neutralized aspirin and the back-titration with the 0.100 M HCl is to confirm the moles of aspirin in the recrystallized aspirin. <em>FALSE. </em>NaOH can be added directly unyil neutralization point because, initially, aspirin can't be dissolved completely
E) Two of the above statements are true. <em>TRUE</em>
<em></em>
Right option is:
<h3>E) Two of the above statements are true.</h3>
Answer: The pH at equivalence point for the given solution is 5.59.
Explanation:
At the equivalence point,
So, first we will calculate the moles of 
as follows.
= 0.0845 mol
Now, volume of 
present will be calculated as follows.
Volume = 
= 
= 0.1891 L
Therefore, the total volume will be the sum of the given volumes as follows.
110 ml + 189.1 ml
= 299.13 ml
or, = 0.2991 L
Now, ![[CH_{3}NH_{3}^{+}] = \frac{0.0845 mol}{0.2991 L}](https://tex.z-dn.net/?f=%5BCH_%7B3%7DNH_%7B3%7D%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B0.0845%20mol%7D%7B0.2991%20L%7D)
= 0.283 M
Chemical equation for this reaction is as follows.
As, 
= 
= 
Now, ![[HNO_{3}] = \sqrt{k_{a}[CH_{3}NH_{3}^{+}]}](https://tex.z-dn.net/?f=%5BHNO_%7B3%7D%5D%20%3D%20%5Csqrt%7Bk_%7Ba%7D%5BCH_%7B3%7DNH_%7B3%7D%5E%7B%2B%7D%5D%7D)
= 
= 
Now, pH will be calculated as follows.
pH = ![-log [H_{3}O^{+}]](https://tex.z-dn.net/?f=-log%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D)
= 
= 5.59
Thus, we can conclude that pH at equivalence point for the given solution is 5.59.
