Rank the following acids from weakest to strongest, based on the following list

Question

Rank the following acids from weakest to strongest, based on the following list of their Ka values 25 degree C. Calculate the pKa values for each acid. HA 3.9 times 10^-2 HB 8.4 times 10^-5 HC 4.7 times l0^-7 HD 2.6 times 10^-4 An acid, HA, has a pKa = 4.7. Calculate the ratio [A]/[HA] for the following pH values: Describe the preparation of 2.00 L of 0.100 M glycine buffer, pH 9.0. from glycine and 0.100 M NaOH. What mass of glycine is required and what volume of NaOH is required? The appropriate pK_a of glycine is 9.6. Calculate the pH of normal arterial blood using the information given below: pK' = 6.1 for CO_2 doubleheadarrow H^+ + HC0_3- [CO_2 + H_2O doubleheadarrow H_2C0_3 doubleheadarrow H^+ + HC0_3 - with pKa = 6.35 for second equilibrium. When both K_eq values are combined the resulting pK' = 6.1 Use this value answer this question so [HA] = [CO_2] and [A] - [HCO_3^-] in the buffering system in the blood. In arterial blood, [HC0_3-] = 24 mM; [C0_2) = 1.2 mM Suppose the normal arterial blood is suddenly made to accept 9 mM of HCl_(aq) i.e. H^+ ions What is the resulting pH of the blood if no CO_2 is allowed to escape? What is the resulting pH of the blood if 9 mM of CO_2 can be quickly exhaled by normal processes? Under these conditions, a person would actually begin to hyperventilate. How will this help to restore normal pH? The R. J. Reynolds Tobacco Company was sued by the State of Minnesota with fraudulently manipulating the pH of cigarette smoke to increase its "kick" i.e. the concentration of nicotine in the bloodstream. They added ammonia to raise the pH from the natural pH of 5.8 to 6.4 Why does the concentration of nicotine in the bloodstream increase with pH of the cigarette smoke. Nicotine is shown below in the fully protonated form at pH 2.5 Show what happens to the molecule when the pH is raised to 5.8. What changes when the pH is raised to 6.4? What pH range would give the maximal absorption of nicotine?

Explanation / Answer

Question 2.

Table:

pH   2.7 3.7   4.7   5.7    6.7   13.7

RATIO ?

Note that this is a buffer, so we can model it with the Henderson Hasselbalch equation

HA <-> H+ + A-

and if we added base, some H+ reacts with OH- to form H2O... but A- ions are left

so there is a common ions effect, A- will help the reaction to form a buffer

if we add H+ ions, the shift goes toward:

H+ + A- --> HA

if we add OH- ions the shift goes:

H+ + OH- ---> H2O

therefore; it is a good buffering system

The Henderson HAsselbalch equation:

pH = pKa + log([Conjugate base]/[Acid])

in this case:

pH = 4.7+ log([A]/[HA])

now...

let us substitute each pH value

pH = 2.7

pH = 4.7+ log([A]/[HA])

2.7 = 4.7 = log([A]/[HA])

10^(2.7-4.7) = [A]/[HA]

[A]/[HA] = 0.01

pH =3.7

3.7 = 4.7 = log([A]/[HA])

10^(3.7-4.7) = [A]/[HA]

[A]/[HA] = 0.1

pH =4.7

4.7= 4.7+ log([A]/[HA])

10^(4.7-4.7) = [A]/[HA]

[A]/[HA] = 1 (same amounts)

pH =5.7

5.7= 4.7+ log([A]/[HA])

10^(5.7-4.7) = [A]/[HA]

[A]/[HA] = 10

pH =6.7

6.7= 4.7+ log([A]/[HA])

10^(6.7-4.7) = [A]/[HA]

[A]/[HA] = 100

pH =13.7

note that buffer is partially destroyed... it is already basic media, meaning H+ ions are pretty inexistent (10^-13 approx)

so

13.7= 4.7+ log([A]/[HA])

10^(4.7-4.7) = [A]/[HA]

[A]/[HA] = 10^-9

or.. HA is almost not dissociated... that is, almost no H+ ions from dissociation

For further questions ,Please consider posting multiple questions in multiple set of Q&A.

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