Proton transfer and Calculate Atom Economy and Reaction Efficiency need help wit

Question

Proton transfer and Calculate Atom Economy and Reaction Efficiency

need help with 2. and 3a.

3a.

Calculate Atom Economy and Reaction Efficiency:

Other info...

Any help at all thanks!!!!

Label the stronger acid, stronger base, weaker acid, and weaker base in Equation 1 for the proton transfer reaction in part A of the experiment Based on information in the "Chemistry and the Environment" section, calculate the atom economy and reaction efficiency of the reaction you carried out in part A. (b) After reading Understanding the Experiment This experiment will help you become familiar with the laboratory environment and with some fundamental laboratory operations, such as measuring mass and volume, separating solids from liquids by vacuum filtration, and drying solids. In part A, you will add hydrochloric acid to a solution of sodium benzoate and, if a different substance forms, recover it and measure its mass. This is a "green" reaction because the only solvent used is water and the only by-product is nontoxic sodium chloride. Although concentrated hydrochloric acid is quite hazardous, the dilute solution of HC1 used here is relatively nontoxic and safe to handle. To understand what is happening in the reaction, you need to know something about the properties of the substances involved and remember what you have previously learned about acid-base chemistry and stoichiometry. When sodium benzoate dissolves in water, it dissociates into benzoate ions (which are weakly basic) and sodium ions, according to the reaction equation: C6H5COONa rightarrowC6H5C00-+Na+ Hydrochloric acid is a solution of hydrogen chloride (HC1) in water. The strongest acid present in this solution is the hydronium ion, formed by the transfer of a proton from an HQ molecule to a water molecule. HCI + H20rightarrow Cl- + H30+ hydronium ion If the hydronium ion concentration is high enough, protons will be transferred from the strong acid H30+ to the basic benzoate ions. This will yield benzoic acid, which, being quite insoluble in water (sodium benzoate is about 200 times more soluble), should precipitate from solution. The reaction is C6H5C00- + H30+ rightarrow C6H5COOH + H20 benzoic acid The net reaction is the sum of these three reaction steps. Key Concept: In a Lowry-Brousted acid-base reaction under standard conditions, the stronger acid transfers protons to a stronger base to yield a weaker acid and a weaker base. C6H5COONa + HCl-rightarrow C6H5COOH + NaCl The hydronium ion concentration depends on the pH of the solution; the lower the pH, the higher [H3O+] will be. By carrying out the experiment, you will discover whether or not the pi I of stomach acid is low enough to cause the conversion of sodium benzoate to benzoic acid. In any experiment that involves the conversion of one substance to another. it is important to understand the stoichiometry of the reaction. (Refer to the appendix "Calculations for Organic Synthesis" if you need a review of stoichiometric calculations.) Equation 2 shows that, if the reaction were complete at used, a mole of benzoic acid would be formed for each mole of sodium benzoate in the reaction mixture. You will start with about 0.400 g of sodium benzoate, which is 2.78 mmol, so the theoretical yield of benzoic acid (if any forms) should be 2.78 mmol as well, corresponding to a

Explanation / Answer

C6H5COOH(aq) + H2O(l) ' H3O+
(aq) + C6H5COO-
(aq)
0.25 - x x x
Ka = 6.3 x 10-5 = ( )( )
( )
3 65
6 5
H O C H COO
C H COOH
+ ?
Ka = 6.3 x 10-5 = ( )( )
( )
x x
0.25 x ? Assume x is small compared to 0.25.
Ka = 6.3 x 10-5 = ( )( )
( )
x x
0.25
x = 3.9686 x 10-3 (unrounded)
Check assumption: (3.9686 x 10-3 / 0.25) x 100% = 2% error, so the assumption is valid.
Percent C6H5COOH Dissociated = Dissociated Acid x 100%
Initial Acid
Percent C6H5COOH Dissociated = x x 100%
0.25 =
3 3.9686 x 10 x 100%
0.25
?
= 1.58745 = 1.6%

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.