ample Test 3B MATCHING (5 pts) 36-40. Match the description to the pH 36. Normal

Question

ample Test 3B MATCHING (5 pts) 36-40. Match the description to the pH 36. Normal blood 37. pH of your blood if you hold your breath 38. pH of your blood if you hyperventilate (breath too fast) 39. pH of a solution of 0.1M acetic acid (pKa 4.74) and 0.1 M sodium acetate 40 pH of 10 mL of the solution in #39 after10mLof0.1g of NaCl has been added 41-45 Choose 41. Reduction of 42. h A) 4.74 B) 7.0 C) 7.2 )7.4 E) 7.6 the product to match the reaction ydration of 43, hydrogenation of 44. Oxidation of 45. Decarboxylation of

Explanation / Answer

Ans. #36. pH of normal blood = D. 7.4

#37. CO2(g) + H2O(l) <--------> H2CO3(aq) <--------> H+ + HCO3-(aq)

The pH of blood is regulated by bicarbonate buffer formed by the catalysis of carbonic anhydrase. Following Le Chatelier’s principle, increase in [CO2] (when holding breath) shifts the equilibrium to the right. That is, holding the breath increases [H+], thus lowers the pH to some extent (because the change is being counter-acted by the buffer system).

So, correct option is- C. 7.2

#38. Under case of hyperventilation, [CO2] is reduced because its exhaled at faster rate. Lower [CO2] thus shifts the equilibrium to the left following Le Chatelier’s principle. Therefore, hyperventilation increase the pH above 7.4.

So, correct option is- E. 7.6

#39. Using HH equation for base-

pH = pKa + log ([A-] / [AH])

Where,            AH = weak acid, A- = Conjugate acid

Now,

            pH = 4.74 +log (0.10/ 0.10) = 4.74 + log 1 = 4.74 + 0 = 4.74

So, correct option is- A. 4.74

#40. Na+ is the cation of a strong base (NaOH). So, it exists in Na+ form and acts as spectator ion only.

#Cl- is the anion of a strong acid (HCl). So, it exists in Cl- form and acts as spectator ion only.

# Therefore, addition of NaCl does not affect the pH of the buffer.

# Since dilution lowers the concertation of weak acid and its conjugate base by same factor, the pH also remains unaffected.

For example, addition of equal volume of NaCl solution halves the concertation of weak acid and its conjugate base.

Now,

            pH = 4.74 +log (0.20/ 0.20) = 4.74 + log 1 = 4.74 + 0 = 4.74

That is, as mentioned above, pH remains unaffected.

So, correct option is- A. 4.74

# Note: Addition of NaOH however, may increase the pH from 4.74 to 7.0

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