Whats the "buffered " bit?
OK! Don't forget you askedA buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit. The buffer range is the pH range where a buffer effectively neutralizes added acids and bases, while maintaining a relatively constant pH.
1. Introduction
2. Example
3. Example 2
4. Buffers in the Human Body
5. References
6. Contributors
Introduction
The equation for pH also shows why pH does not change by much in buffers.
Ka=[H+][A−][HA]
pH=pKa+log[A−][HA]
Where,
A− is the concentration of the conjugate base
HA is the concentration of the acid
When the ratio between the conjugate base/ acid is equal to 1, the pH = pKa. If the ratio between the two is 0.10, the pH drops by 1 unit from pKa since log (0.10) = -1. If a ratio increases to a value of 10, then the pH increases by 1 unit since log (10) = 1. The buffer capacity has a range of about 2. This means when a buffer is created, the pH can be changed by -1 by acid or +1 by base before the pH begins to change substantially. After the addition of base to raise the pH by 1 or more, most of the conjugate acid will have been depleted to try to maintain a certain pH, so the pH will be free to increase faster without the restraint of the conjugate acid. The same goes for the addition of acid, once the conjugate base has been used up, the pH will drop faster since most of the conjugate base has been used up.
Weak Acid.jpg
Example
What is the effect on the pH of adding 0.006 mol HCL to 0.3L of a buffer solution that is 0.250M HC2H3O2 and 0.560 M NaC2H3O2? pKa= 4.74
pH=4.74+log0.5600.250=4.74+0.35=5.09
C2H3O−2+H3O+⇌HC2H3O2+H2O
Calculate the starting amount of C2H3O2-
0.300L×0.560M=0.168molC2H3O−2
Calculate the starting amount of HC2H3O2
0.300L×0.250M=0.075molHC2H3O2
C2H3O2- H3O- HC2H3O2
Original Buffer 0.168 mol 0.075 mol
Add 0.006 mol
Change -0.006 mol -0.006 mol +0.006 mol
Final Amount 0.162 mol 0.081 mol
Now calculate the new concentrations of C2H3O2- and HC2H3O2:
0.162mol0.300L=0.540MC2H3O−2
0.081mol0.300L=0.540MHC2H3O2
Using the new concentrations, we can calculate the new pH:
pH=4.74+log0.5400.270=4.74+0.30=5.04
Calculate the pH change:
pHfinal−pHinitial=5.04−5.09=−0.05
Therefore, the pH dropped by 0.05 pH units.
Example 2
What is the effect on the pH of adding 0.006 mol NaOH to 0.3L of a buffer solution that is 0.250M HC2H3O2 and 0.560 M NaC2H3O2? pKa= 4.74
pH=4.74+log0.5600.250=4.74+0.35=5.09
HC2H3O2+OH−⇌C2H3O−2+H2O
Calculate the starting amount of HC2H3O2
0.300L×0.250M=0.075molHC2H3O2
Calculate the starting amount of C2H3O2-
0.300L×0.560M=0.168molC2H3O−2
HC2H3O2 OH- C2H3O2-
Original Buffer 0.075 mol 0.168 mol
Add 0.006 mol
Change -0.006 mol -0.006 mol +0.006 mol
Final Amount 0.069 mol 0.174 mol
Now calculate the new concentrations of HC2H3O2 and C2H3O2-:
0.069mol0.300L=0.230MHC2H3O2
0.174mol0.300L=0.580MC2H3O−2
Using the new concentrations, we can calculate the new pH:
pH=4.74+log0.5800.230=4.74+0.40=5.14
Calculate the pH change:
pHfinal−pHinitial=5.14−5.09=+0.05
Therefore, the pH increased by 0.05 pH units.
GOD I'M GOOD DID YOU GET ALL THAT 