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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 2.50 L of a 0.500 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.
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Concepts and reason

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.

Fundamentals

Henderson-Hasselbalch equation is a mathematical expression which can be used to calculate the of buffer solutions.

[Salt]
pH = PK + log
[Acid]

K,, is thedissociation constant of a weak acid.
[Acid]= Concentration of a weak acid.
[Salt]= Concentration of its salt.

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

Write the Henderson-Hasselbalch equation of basic buffer calculation.

pOH =pK, +log;
[base]

K,, is thedissociation constant of a weak base.
[base]= Concentration of a weak base.
[Salt]= Concentration of its salt.

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

The reaction for given buffer solution can be written as follows

NH,Cl
+NH, +HCl

Given data:

pH of buffer = 8.74
K, ammonia = 1.8x10-5

So,

pH+pOH = 14
pOH =14-PH
pOH =14-8.74
pOH = 5.26

Then of given solution can be calculated as follows

pK, =-log(K)
pKg =-log(1.8x10)
PK, = 4.74

Henderson-Hasselbalch was used for calculating the mass of NH,CI
as given below, the values of pOH and pk
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,]

Then

3.31- (NH,CI]
[NH.]
since[NH,]=0.500M

Finally

[NH,Cl]=3.31x0.500M
[NH,Cl] =1.655M

The mole of ammonium chloride was calculated as follows:

Mole of NH,CI=1.655mol/L 2.50L
= 4.1375 mol

The mass of ammonium chloride is calculated as follows:

Mass of NH CI=4.1375mol x 53.491g/mol (molar mass of NH,CI= 53.491g)
= 221g

Ans:

The mass of NH,CI
is.

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