pHof blood plasma is 7.40. Assuming the principal buffer system is HCo calculate

Question

pHof blood plasma is 7.40. Assuming the principal buffer system is HCo calculate the ratio [Hco, WH CosJ this buffer more effective against an added or added base? pha a calculate the pH of the 0.20 MNH30 20 MNHa buffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10 MHCl to 65.0 mL of the buffer? 13. The pKs of two monoprotic acids HA and HB are 5.9 and 8.1. respectively. Which of the two is the stronger acid? 14. A diprotic acid, H2A, has the following ionization K.2- 2.5 x 100. In order to p a solution x 103 would you choose? of pH 0, which combination a. NaHA/H24. b, Na2A/NaHA.

Explanation / Answer

11) pH of blood plasma = 7.40, [HCO3-]/[H2CO3] = ? pKa = 6.37 for .H2CO3.

Formula, pKa = pH + log[base/acid].............for any Buffer

so, pH = pKa + log{[HCO3-]/[H2CO3]}.....................For HCO3-/H2CO3 buffer

Let us put all the known values,

7.40 = 6.37 + log{[HCO3-]/[H2CO3]}

so, log{[HCO3-]/[H2CO3]} = 7.40 - 6.37

log{[HCO3-]/[H2CO3]} = 1.03.............(logarithmic form)

[HCO3-]/[H2CO3] = 101.03.............(exponential form)

[HCO3-]/[H2CO3] = 10.72.

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12) a) pH of 0.20 M NH3 / 0.20 M NH4Cl

NH3 + H2O --------> NH4OH + OH- .

i.e. its a 0.20 M NH4OH / 0.20 M NH4Cl buffer and it is a BASIC buffer.

We have, Kb = 1.8 x10-5 for NH4OH . [NH4Cl] = 0.20 M, [NH4OH] = 0.20 M.

Handerson's equation hence will be,

pOH = -log Kb + log{[salt]/[base]}

pOH = -logKb + log{[NH4Cl/[NH4OH]}.............(I)

pOH = -log(1.8x10-5) + log(0.2/0.2)

pOH = 4.74 + log(1)

pOH = 4.74 + 0

pOH = 4.74

pH+pOH = 14

so, pH + 4.74 = 14

pH = 14 - 4.7

pH = 9.26.

b) Addition of HCl will increase the concentration of NH4Cl and decreases of that of NH4OH

10 mL, 0.1 M HCl = 1 milimole HCl

and 65 mL buffer 0.2 M in NH4OH

so, 65 mL buffer, 0.2 M NH4OH = 65 x 0.2 = 13 milimole

So new concentrations are,

13-1 = 12 milimole NH4OH

13+1 = 14 milimole NH4Cl

use these values in Handerson equation equation,

pH = pKa + log{[NH4OH]/[NH4Cl]}

pH = 9.26 +log[12/14]

pH = 9.26 - 0.067

pH = 9.19.

pH lowered by 0.067 and new pH = 9.19.

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13) pKa and pH are equivalent terms. We know that, smaller the pH value stronger is the acid hence simmilarly lower the pKa stronger is the acid.

pKa of HA =5.9 and pKa of HB= 8.1

pKa of HA < pKa of HB

so HA is stronger acid than HB.

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