tion of KOH and 13) E 13) The to water produces a buffer solution. A) HI B) NH3
ID: 304492 • Letter: T
Question
tion of KOH and 13) E 13) The to water produces a buffer solution. A) HI B) NH3 C) LiC2H302 D) KF none of the above 14) Which of the following could be added to a solution of sodium acetate to produce a buffer? 14) cetic acid or hydrochloric acid hydrochloric acid only C) potassium acetate only E) acetic acid only 15) What is the primary buffer system that controls the pH of the blood? 15) t A) carbonate, bicarbonate B) carbonate, carbonic acid C) carbonic acid, carbon dioxide D) carbon dioxide, carbonate E carbonic acid, bicarbonate 12.0 PHof 100 solution 80 in fask 6.0 Equivalence Point 4.0 2.0 5 10 15 20 25 30 35 40 45 mL of 0.115 M NaOH added to fask 16) A 25.0 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH 16) solution. The titration curve above was obtained. The unknown compound is A) a strong base B) a weak base a strong acid a weak acid neither an acid nor a base 17) A 50.0 mL sample of a solution of a monoprotic acid is titrated with a 0.115 M NaOH solution. The 17) titration curve above was obtained. The concentration of the monoprotic acid is about mol/L. 0.0600 B) 0.120 C) 0.100 D) 25.0Explanation / Answer
13) An acidic buffer consists of a weak acid and its salt with a strong base, for example, acetic acid/sodium acetate buffer. A basic buffer consists of a weak base and its salt with a strong acid, for example, ammonia/ammonium chloride buffer. KOH is a strong base and hence, cannot form a buffer by itself. KOH may be added to a solution of a weak acid to produce a buffer. HI is a strong acid while NH3 is a weak base. LiC2H3O2 and KF are salts. Thus, none of the options given in (A) – (D) can form a buffer with KOH. Thus, the correct answer is
(E) none of the above.
14) Sodium acetate (NaC2H3O2) is the salt of a weak acid (acetic acid, HC2H3O2) and a strong base (sodium hydroxide, NaOH). Sodium acetate can form a buffer with its conjugate weak acid, i.e, acetic acid only. Hence, if acetic acid is added to the solution of sodium acetate, a buffer solution will result. Alternatively, sodium acetate (completely ionized in aqueous solution) can be partly neutralized when hydrochloric acid (HCl) is added to sodium acetate. The resulting solution will contain un-neutralized sodium acetate, acetic acid and sodium chloride. Sodium chloride has no buffering action.
NaC2H3O2 (aq) + HCl (aq) ----------> HC2H3O2 (aq) + NaCl (aq)
Thus, the correct answer is
(A) acetic acid or hydrochloric acid.
15) The pH of blood is maintained in the body by the buffering action of carbonic acid (H2CO3) and bicarbonate (HCO3-). The blood buffer is H2CO3/HCO3- and controls the pH of the body by acid-base homeostatic mechanism. Therefore, the correct answer is (E).
16) The titration curve indicates that a strong base (NaOH) is added to a solution of a chemical species which can participate in acid-base titration. The chemical species titrated is a weak acid. As the base is run into the acid, the pH rises rapidly. The base reacts with the weak acid to form the salt of the weak acid (which functions as the conjugate base of the acid). Therefore, the system now contains a weak acid and its conjugate base and thus, forms a buffer system. The buffer system resists the change in pH upon further addition of base and thus, the pH increases slowly after a certain point till the equivalence point is reached. Beyond the equivalence point, the solution contains excess strong base and thus, the pH rises sharply. Thus, the correct answer is (D) a weak acid.
17) The titration curve shows that the equivalence point is reached when 25 mL of 0.115 M NaOH is added to the weak acid.
Therefore, moles of NaOH added = (volume of NaOH in L)*(molarity of NaOH) = (25 mL)*(1 L/1000 mL)*(0.115 M) = 0.002875 mole.
At the equivalence point, the moles of NaOH = moles of weak acid neutralized = 0.002875 mole.
The volume of the solution is 50.0 mL = (50.0 mL)*(1 L/1000 mL) = 0.05 mole.
Molarity of the weak acid = (0.002875 mole)/(0.05 L) = 0.0575 M ? 0.0600 M.
The correct answer is (A).
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