Why does buffer capacity increase as a solution becomes very acidic (pH ≈ 1) or very basic (pH ≈ 13)?

Short Answer

Expert verified

At a very low pH, i.e., very acidic (pH ≈ 1) or at a very high pH, i.e., very basic (pH ≈13), the buffer capacity increases as it resists changes in pH by either absorbing or desorbing H+ and OH-ions.

Step by step solution

01

Definition of buffer capacity

Buffer capacity is a quantitative measure of how much a solution is reluctant to pH changes when strong acid or base is added. The more resistant the solution is to pH change, the greater the buffer capacity.

02

Impact on buffer capacity when pH is very acidic or very basic

At very low or very high pH values, there are high acid or base concentrations in a solution. So, adding a little bit of acid or base in that solution would have nearly no effect.

Also, we can consider the following:

  1. At low pH buffer isH3O+/H2O
  2. At high pH buffer is H2O/OH-
03

Conclusion

Buffer capacity increases as a solution become very acidic or very basic. A buffer is most effective in resisting changes in pH when pH ≈ 1 or ≈ 13. Buffer capacity increases at high pH (and at low pH), simply because there is a high concentration of OH- at high pH (and H+ at low pH). Hence, adding a small amount of acid or base to a large amount of OH- (or H+) will not have a large effect on pH.

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Most popular questions from this chapter

(a) Calculate the pH of a solution prepared by mixing 0.00100molof the baseB(Kb=10-2.00)with 0.0200mol of BH+Br-and diluting to 1.00L.First calculate the pH by assuming[B]=0.0100and [BH+]=0.0200M . Compare this answer with the pH calculated without making such an assumption.

(b) After working part (a) by hand, use Excel Goal Seck to find the same answers.

Select a compound from Table 9-2 that you could use to make 250mLof 0.2Mbuffer with a pH of 6.0. Explain how you would make the buffer.

Which of the following bases would be most suitable for preparing a buffer of pH9.00?

(i) NH3(ammonia,Kb=1.76×10-5);

(ii) role="math" localid="1654764490555" C6H5NH2(aniline,role="math" localid="1654764515634" C6H5NH2 );

(iii)H2NNH2(hydrazine,Kb=1.05×10-6)));

(iv) C6HsNH2(pyridine,Kb=1.58×10-9) ).

Calculate the pH of a solution prepared by mixing 0.0800mol of chloroacetic acid plus0.0400mol of sodium chloroacetate in1.00L of water.

(a) First do the calculation by assuming that the concentrations of HA andA- equal their formal concentrations.

(b) Then do the calculation, using the real values of and in the solution.

(c) Using first your head, and then the Henderson-Hasselbalch equation, find the pH of a solution prepared by dissolving all the following compounds in one beaker containing a total volume of1.00L:0.180molCICHCOO2H,0.020molCICHCHO2Na20.080molHNO3,and0.080molCa(OH)2. Assume thatCa(OH)2 dissociates completely.

What is the pH of a solution prepared by dissolving 1.23gof

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