Buffer:
A buffer is a chemical solution that contains a weak acid/base mixed with its conjugate base/acid (as in the form of salts). It can resist
change when a small amount of a strong acid/base is added to it. The acidic buffer solution is made up of less than
and the alkali buffer solution is made up of more than
.
Henderson-Hasselbalch equation is a mathematical expression which can be used to calculate the
of buffer solutions.
![[Salt]
pH = PK + log
[Acid]](http://img.homeworklib.com/questions/5ec845f0-0d23-11ea-a772-67f967f30e43.png?x-oss-process=image/resize,w_560)
![K,, is thedissociation constant of a weak acid.
[Acid]= Concentration of a weak acid.
[Salt]= Concentration of its salt.](http://img.homeworklib.com/questions/5f269a80-0d23-11ea-b9c8-23a5fc023993.png?x-oss-process=image/resize,w_560)
Concentration of acid and base can be calculated as follows:
![number of moles of weak acid
volume of solvent in liters
_ number of moles of salt
[Salt] =
volume of solvent in liters](http://img.homeworklib.com/questions/5f65aba0-0d23-11ea-9885-0dec9560ad03.png?x-oss-process=image/resize,w_560)
Write the Henderson-Hasselbalch equation of basic buffer calculation.
![pOH =pK, +log;
[base]](http://img.homeworklib.com/questions/5fa91a00-0d23-11ea-963a-1799859d62c9.png?x-oss-process=image/resize,w_560)
![K,, is thedissociation constant of a weak base.
[base]= Concentration of a weak base.
[Salt]= Concentration of its salt.](http://img.homeworklib.com/questions/5fec2a10-0d23-11ea-85c3-37b70d027934.png?x-oss-process=image/resize,w_560)
Concentration of acid and base can be calculated as follows:
![1_ number of moles of weak base
volume of solvent in liters
_ number of moles of salt
[Salt] =
volume of solvent in liters](http://img.homeworklib.com/questions/604c1580-0d23-11ea-a720-174a70b51371.png?x-oss-process=image/resize,w_560)
The reaction for given buffer solution can be written as follows

Given data:

So,

Then
of given solution can be calculated as follows

Henderson-Hasselbalch was used for calculating the mass of
as given below, the values of
were substituted in Henderson-Hasselbalch equation.
![NH,CI]
5.26=4.74+log
[NH,
5.26-4.74 = log INH,CI]
* [NH,].
0.52 = log (NH,CI]
[NH ]
10952 _ [NH,CI]
[NH,]](http://img.homeworklib.com/questions/630625d0-0d23-11ea-9264-296b6b88ebfa.png?x-oss-process=image/resize,w_560)
Then
![3.31- (NH,CI]
[NH.]
since[NH,]=0.500M](http://img.homeworklib.com/questions/63495830-0d23-11ea-9e58-a3f750507761.png?x-oss-process=image/resize,w_560)
Finally
![[NH,Cl]=3.31x0.500M
[NH,Cl] =1.655M](http://img.homeworklib.com/questions/63aad660-0d23-11ea-acab-677e0eef12a5.png?x-oss-process=image/resize,w_560)
The mole of ammonium chloride was calculated as follows:

The mass of ammonium chloride is calculated as follows:

The mass of
is
.
How many grams of dry NH4Cl need to be added to 2.50 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer soluti...
How many grams of dry NH4Cl need to be added to 1.50 L of a 0.500 M solution of ammonia, NH3,to prepare a buffer solution that has a pH of 8.79? Kb for ammonia is 1.8*10^-5.
How many grams of dry NH4Cl need to be added to 2.50 L of a 0.400 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.67? Kb for ammonia is 1.8×10−5.
How many grams of dry NH4Cl need to be added to 2.10 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.76? Kb for ammonia is 1.8×10−5. Express your answer with the appropriate units.
How many grams of dry NH4Cl need to be added to 2.30 L of a 0.100 M solution of ammonia,NH3, to prepare a buffer solution that has a pH of 9.00? Kb for ammonia is 1.8x 10^-5.
How many grams of dry NH4Cl need to be added to 1.70 L of a 0.100 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.62? Kb for ammonia is 1.8×10−5.
How many grams of dry NH4Cl need to be added to 1.70 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.80? Kb for ammonia is 1.8×10−5.
How many grams of dry NH4Cl need to be added to 2.40 L of a 0.400 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.78? Kb for ammonia is 1.8 x 10^-5.
How many grams of dry NH4Cl need to be added to 1.70 L of a 0.500 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.51? Kb for ammonia is 1.8 x 10^-5.
How many grams of dry NH4ClNH4Cl need to be added to 2.50 LL of a 0.800 MM solution of ammonia, NH3NH3, to prepare a buffer solution that has a pHpH of 8.86? KbKbK_b for ammonia is 1.8×10−51.8×10−5.
How many grams of dry NH4Cl need to be added to 1.90 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.74? Kb for ammonia is 1.8×10^-5.