ANSWER D,E & FIn some natural systems the pH must be maintained within very narrow limits. e.g. in human blood the pH must remain close to 7.4 or cell deterioration occurs. Blood contains several weak acid/conjugate base equilibria called buffers, which control the pH.One weak acid present in blood is the dihydrogen phosphate ion, H2PO4-(aq), for which the equilibrium in aqueous solution is H2PO4-(aq) +H2O(l) <--> HPO42-(aq) + H3O+(aq).a) What would be the effect on this equillibrium of adding hydrochloric acid solution?b) Write the expression for the equilibrium constant of this reaction.c) If 0.5 mol H2PO4- and 0.5 mol HPO42- are in equilibrium in 1.0L of aqueous solution, calculate the pH of the solution. (Ka for H2PO4-= 6.4 x 10-8)d) 0.010 mol HCl is now added to the 1.0L of the solution in c). Assuming that all the added H+ ions are used up in the equilibrium shift, calculate the concentrations of H2PO4- and HPO42-. Hence calcuate the pH of this solution.e) 0.010 mol HCl has been added to water then made up to 1.0L of solution. Calculate the pH.f) From the above results comment on the effectiveness of the H2PO4- / HPO42- equilibrium on the control of pH in blood.

Question
Asked Oct 23, 2019

ANSWER D,E & F

In some natural systems the pH must be maintained within very narrow limits. e.g. in human blood the pH must remain close to 7.4 or cell deterioration occurs. Blood contains several weak acid/conjugate base equilibria called buffers, which control the pH.

One weak acid present in blood is the dihydrogen phosphate ion, H2PO4-(aq), for which the equilibrium in aqueous solution is H2PO4-(aq) +H2O(l) <--> HPO42-(aq) + H3O+(aq).

a) What would be the effect on this equillibrium of adding hydrochloric acid solution?

b) Write the expression for the equilibrium constant of this reaction.

c) If 0.5 mol H2PO4- and 0.5 mol HPO42- are in equilibrium in 1.0L of aqueous solution, calculate the pH of the solution. (Ka for H2PO4-= 6.4 x 10-8)

d) 0.010 mol HCl is now added to the 1.0L of the solution in c). Assuming that all the added H+ ions are used up in the equilibrium shift, calculate the concentrations of H2PO4- and HPO42-. Hence calcuate the pH of this solution.

e) 0.010 mol HCl has been added to water then made up to 1.0L of solution. Calculate the pH.

f) From the above results comment on the effectiveness of the H2PO4- / HPO42- equilibrium on the control of pH in blood.

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Expert Answer

Step 1

c) As already calculated in part c, the pH of the solution is 7.2

Now the Henderson equation is

PH=pKa+log conjugate acid
[conjugate base]
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PH=pKa+log conjugate acid [conjugate base]

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Step 2

 When 0.01 mol of HCl is added, we are basically adding 0.01 mol of Hydronium ions. Thus 0.01 mol of conjugate acid is increased and 0.01 mole of conjugate base is decreased.

Therefore,

[conjugate acid]
pH pKa+log
[conjugate base
[0.50.01
[0.5-0.01
= 7.2 log
- 7.2 0.0170
=7.2170
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[conjugate acid] pH pKa+log [conjugate base [0.50.01 [0.5-0.01 = 7.2 log - 7.2 0.0170 =7.2170

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Step 3

the concentration of H2PO4-  is 0.51mol and HPO4...

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