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Chapter 10: Q20P (page 231)

Find the pH and the concentration of each species of lysine in a solution of 0.0100 M lysine? HCl, lysine monohydrochloride. The notation 鈥渓ysine? HCl鈥� refers to a neutral lysine molecule that has takenon one extra proton by addition of one mole of HCl. A more meaning-

ful notation shows the salt (lysineH1)(C12) formed in the reaction..

Short Answer

Expert verified

The PH of the supplied solution is 5.64, and the concentration is2.40脳10-11M

Step by step solution

01

Define the formula to find the concentration of ph.

The formula PH of the solution

[H+]=K2K3F+K2KwK2+F

02

Calculate the concentration of PH .

Lysine hydrochloride that means Lysine has amine group that can be protonated.

The pHof solution

H+=K1K20.0100+K1KwK1+0.0100=2.32脳10-6MnpH=5.64H2L+=0.0100MH3L2+=H+H2L+K1=1.36脳10-6MnHL=K2H2L+K1=3.68脳10-6Mn

L-=K3HLH+=2.40脳10-11M

Thus, the PH of the supplied solution is 5.64, and the concentration is2.40脳10-11M

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Most popular questions from this chapter

The both [H2A]=[HA鈥�]are equal as found to be pH=10.54

Heterogeneous equilibrium.CO2dissolves in water to give "carbonic acid" (which is mostly dissolvedCO2, as described in Box 6-4).

CO2(g)饾啅CO2(aq)K=10-1.5

(The equilibrium constant is called the Henry's law constant for carbon dioxide, because Henry's law states that the solubility of a gas in a liquid is proportional to the pressure of the gas.) The acid dissociation constants listed for "carbonic acid" in Appendix G apply toCO2(aq). Given thatPCO2in the atmosphere is 10-3.4atm, find the pHof water in equilibrium with the atmosphere.

(a) Calculate the quotient in [H3PO4]/[H3PO-4]in0.0500MKH2PH4.

(b) Find the same quotient for 0.0500MK2PO4.

Neutral lysine can be written HL. The other forms of lysine areH3L2+,H2L+,andL-. The isoelectric point is the pH at which the average charge of lysine is 0. Therefore, at the isoelectric point,2[H3L2+]+[H2L+]=[L-]. Use this condition to calculate the isoelectric pH of lysine.

(a) Copy column B of your spreadsheet and paste it into columnG. ChangeK1to10-4in cell G5 and change K2 to 10-8 in cell G6. Change F to 0.01 in cell G8. Column G now contains concentrations for the amphiprotic salt Na+HA-withK1=10-4,K2=10-8, andF=0.01M. Check the answers by hand beginning with pH=12(pK1+pK2). With[HA-],F, calculate[H2A]and[A2-]. Then find[HA-]>>F-[H2A]-[A2-].

(b) Copy column G of your spreadsheet and paste it into column H. Change K2 to 10-5 in cell G6. Column G now contains the concentrations for the intermediate form of a diprotic acid with K1=10-4, K2=10-5, and F =0.01M. You should observe[HA-]=6.13脳10-3Mand pH=4.50.
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