Answer
Verified
409.2k+ views
Hint: A buffer solution is a solution which resists a change in the pH due to the addition of acid or base. The pH of a buffer solution is expressed in terms of the dissociation constant of the acid or and its conjugate base.
Complete step by step answer:
Buffer solution is of two types named as acidic buffer or a basic buffer. An Acid buffer has an acidic \[pH\] which is made of a weak acid and its salt with a strong base. Similarly a basic buffer has a basic \[pH\] made of a weak base and its salt with a strong acid.
The addition of sodium hydroxide to a solution reacts with the acid \[HA\] and produces a salt of sodium as \[N{a^ + }{A^ - }\] and water. The acid and the base \[NaOH\] react in \[1:1\] mole ratios.
\[HA\left( {aq} \right) + O{H^ - }\left( {aq} \right) \to N{a^ + }{A^ - }\left( {aq} \right) + {H_2}O\left( l \right)\]
Next we determine the number of moles of weak acid and the conjugate base present in the buffer before the addition of the strong base. The molarity of the acid is given as \[0.1M\] and the volume of the buffer solution is \[100mL\].
So moles of \[HA\] = moles of \[{A^ - }\] = \[0.1M \times 100mL = \dfrac{{0.1{\text{ }}moles}}{{1000mL}} \times 100mL = 0.01moles\].
Before determining the actual amount in grams of sodium hydroxide we calculate the number of moles of sodium hydroxide added to buffer solution. Let \[x\] be the number of moles of sodium hydroxide added to the buffer.
So at the end of the reaction between \[NaOH\] and \[HA\] , the resulting solution contains \[0.01 - x\] moles of \[HA\] where \[x\] moles of the acid is consumed. The moles of the conjugate base \[{A^ - }\] produced in the reaction is \[0.01 + x\] moles.
The \[pH\] of a buffer solution of weak acid and conjugate base pair is given by Henderson - Hasselbalch equation. It is represented as
\[pH = pKa + log\dfrac{{[A - ]}}{{[HA]}}\]
Given the final \[pH\] of the solution is \[5.5\] and the \[pKa\] of the acid is \[5\].
Thus \[pH = pKa + log\left[ {\dfrac{{\left( {0.01 + x} \right)moles}}{{\left( {0.01 - x} \right)moles}}} \right]\]
\[5.5 = 5 + log\left( {\dfrac{{0.01 + x}}{{0.01 - x}}} \right)\]
\[log\left( {\dfrac{{0.01 + x}}{{0.01 - x}}} \right) = 0.5\]
\[\dfrac{{0.01 + x}}{{0.01 - x}} = anti\log (0.5)\]
\[\dfrac{{0.01 + x}}{{0.01 - x}} = 3.16\]
\[0.01 + x = 0.0316 - 3.16x\]
\[x = \dfrac{{0.0316 - 0.01}}{{1 + 3.16}} = 0.00519moles.\]
The molar mass of sodium hydroxide is \[40g/mol\]. Thus the amount of sodium hydroxide is calculated as
$ = 0.00519moles \times 40g/mol = 0.21g$.
Hence \[0.21g\] of solid \[NaOH\] must be added to a buffer solution to make the pH of solution \[5.5\].
Note: This is an example of a buffer solution of weak acid and its salt. The Henderson Hasselbalch equation also applies for a weak base and salt buffer. Buffer system is very important in several biological processes to maintain the pH of the blood. The enzymatic activity in living systems is dependent on the pH of the blood.
Complete step by step answer:
Buffer solution is of two types named as acidic buffer or a basic buffer. An Acid buffer has an acidic \[pH\] which is made of a weak acid and its salt with a strong base. Similarly a basic buffer has a basic \[pH\] made of a weak base and its salt with a strong acid.
The addition of sodium hydroxide to a solution reacts with the acid \[HA\] and produces a salt of sodium as \[N{a^ + }{A^ - }\] and water. The acid and the base \[NaOH\] react in \[1:1\] mole ratios.
\[HA\left( {aq} \right) + O{H^ - }\left( {aq} \right) \to N{a^ + }{A^ - }\left( {aq} \right) + {H_2}O\left( l \right)\]
Next we determine the number of moles of weak acid and the conjugate base present in the buffer before the addition of the strong base. The molarity of the acid is given as \[0.1M\] and the volume of the buffer solution is \[100mL\].
So moles of \[HA\] = moles of \[{A^ - }\] = \[0.1M \times 100mL = \dfrac{{0.1{\text{ }}moles}}{{1000mL}} \times 100mL = 0.01moles\].
Before determining the actual amount in grams of sodium hydroxide we calculate the number of moles of sodium hydroxide added to buffer solution. Let \[x\] be the number of moles of sodium hydroxide added to the buffer.
So at the end of the reaction between \[NaOH\] and \[HA\] , the resulting solution contains \[0.01 - x\] moles of \[HA\] where \[x\] moles of the acid is consumed. The moles of the conjugate base \[{A^ - }\] produced in the reaction is \[0.01 + x\] moles.
The \[pH\] of a buffer solution of weak acid and conjugate base pair is given by Henderson - Hasselbalch equation. It is represented as
\[pH = pKa + log\dfrac{{[A - ]}}{{[HA]}}\]
Given the final \[pH\] of the solution is \[5.5\] and the \[pKa\] of the acid is \[5\].
Thus \[pH = pKa + log\left[ {\dfrac{{\left( {0.01 + x} \right)moles}}{{\left( {0.01 - x} \right)moles}}} \right]\]
\[5.5 = 5 + log\left( {\dfrac{{0.01 + x}}{{0.01 - x}}} \right)\]
\[log\left( {\dfrac{{0.01 + x}}{{0.01 - x}}} \right) = 0.5\]
\[\dfrac{{0.01 + x}}{{0.01 - x}} = anti\log (0.5)\]
\[\dfrac{{0.01 + x}}{{0.01 - x}} = 3.16\]
\[0.01 + x = 0.0316 - 3.16x\]
\[x = \dfrac{{0.0316 - 0.01}}{{1 + 3.16}} = 0.00519moles.\]
The molar mass of sodium hydroxide is \[40g/mol\]. Thus the amount of sodium hydroxide is calculated as
$ = 0.00519moles \times 40g/mol = 0.21g$.
Hence \[0.21g\] of solid \[NaOH\] must be added to a buffer solution to make the pH of solution \[5.5\].
Note: This is an example of a buffer solution of weak acid and its salt. The Henderson Hasselbalch equation also applies for a weak base and salt buffer. Buffer system is very important in several biological processes to maintain the pH of the blood. The enzymatic activity in living systems is dependent on the pH of the blood.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE