When there are equal number of H+ and OH- ions, the pH of water is 7.
Answer and Explanation:
<em>A funnel is in the top of the buret and a beaker is positioned underneath the buret:</em> This is correct and is necessary to fill the buret, but the funnel and the beaker has to be removed before the titration starts. The calculation for moles of analyte does not affect.
<em>A solution is being poured from a bottle into the buret via the funnel:</em> Using a funnel helps to fill the burette but it must be removed to filling the buret at 0.0 mL. In this case, the calculation for moles of analyte do not affect.
<em>Adding titrant past the color change of the analyte solution</em>: In this case, an excess of titrant is added, thus the calculation for moles of anality will be higher than it should be.
<em>Recording the molarity of titrant as 0.1 M rather than its actual value of 0.01 M</em>: In this case, the titrant is considered more concentrated than it is hence, the calculation for moles of anality will be higher than it should be.
<em>Spilling some analyte out of the flask during the titration</em>: The excess of titrant spilled out of the flask higher up the volume of titrant measured. Therefore, the calculation for moles of anality will be higher than it should be.
<em>Starting the titration with air bubbles in the buret</em>: The air inside the burette occupies measured volume, thus the volume of titrant measured will be higher than the real volume spilled in the flask. Hence the calculation for moles of anality will be higher than it should be.
<em>Filling the buret above the 0.0 mL volume mark</em>: Some volume of titrant will be spilled inside the flask but will no be measured since the buret measures the titrant below the 0.0mL mark, thus the calculation for moles of anality will be lower than it should be.
Answer: p2 = 1.06p1
Explanation: pressure increases with temperature increase.
According to Gass law
P1/T1 = P2/T2
T1 = 20°c = 20 +273 = 293k
T2 = 40°c = 40 +373 = 313k
Therefore
P2 = P1T2/T1 = 313P2/293
P2 = 1.06P1
Answer:
(A) pH < 1 the predominant form is the cation: H3C-C(H)(NH3+)-COOH
(B) pH = pl the predominant form is the zwitterion H3C-C(H)(NH3+)-COO-
(C) pH > 11 the predominant form is the anion: H3C-C(H)(NH2)-COO-
(D) Does not occurs in any significant pH: H3C-C(H)(NH2)-COOH
Explanation:
Amino acids are bifunctional because they have an amine group and a carboxyl group. The amine group is a weak base and the carboxyl group is a weak acid, but the pKa of both groups will depend on the whole structure of the amino acid. Also, every amino acid has an isoelectric point (pI), which means the pH were the predominant form of the amino acid is the zwitterion. The structure of the alanine (CH3CH2NH2COOH) shows it has the carboxyl group at C1 with a pKa1 of 2.3 and the amino group at C2 whit the pKa2 of 9.7. The isoelectric poin (pI) of Alanine is 6. Consequently, the protonation of the molecule will depend on the pH of the solution. There are three possibilities:
1) If the pH is under the pKa of the carboxyl group (2.3) the predominant form will be with the amino group protonated, forming a cation (CH3CH(NH3+)COOH).
2) If the pH is between pKa1 (2.3) and pKa2 (9.7) the predominant form will be the zwitterion (CH3CH(NH3+)(COO-)).
3) If the pH is upper the pKa2 of the amino group (9.7) the predominant form will be with the carboxyl group deprotonated, forming an anion (CH3CHNH2(COO-)).