Please provide steps by step to solution. There is no additional information, th

Question

Please provide steps by step to solution. There is no additional information, that is how it is shown in a previous exam.

The cytoplasm of cells contains many compounds which are weak acids and which are therefore partially dissociated at the cell’s pH. In this respect the cytoplasm is a multicomponent buffer. Consider a two-

component buffer of the acids HX and HY, which can dissociate to H+, X- and Y-. Their pKa values are 6.8 (HX) and 7.4 (HY). The pH is 7.2. The solution is prepared to be 0.1 M in HX and 0.1 M in HY,

i.e. [HX ] + [X-] = 0.1 M and similarly for HY.

(a) Set up BUT DO NOT SOLVE the expressions needed to determine the concentrations of the undissociated acids and of their anions, explaining in words the logic you used to set up the calculation.

b) Solve for the concentrations of HX, HY, X- and Y-. PART (b) MUST BE SEPARATED FROM PART (a)

Explanation / Answer

(a) pH = pKa + log([X-]/[HX]

with,

pH = 7.2

pKa = 6.8

[HX] + [X-] = 0.1 M

[X-] = 0.1 - [HX]

we get,

7.2 = 6.8 + log(0.1- [HX]/[HX])

Similarly for HY,

7.2 = 7.4 + log([0.1 - [HY]/[HY])

(b) For HX

7.2 = 6.8 + log([0.1 - [HX]/[HX])

2.512[HX] = 0.1 - [HX]

[HX] = 0.0285 M

[X-] = 0.0715 M

For HY,

7.2 = 7.4 + log([0.1 - [HY]/[HY])

0.631[HY] = 0.1 - [HY]

[HY] = 0.0613 M

[Y-] = 0.0387 M

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