Answer
Verified
385.5k+ views
Hint: We have the concentration of salt and base, and $p{K_a}$ we will use this value to find pK$_b$ then put it in the formula for basic buffer pOH = pK$_b$ +log$\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$ and then use the value of pOH to find the pH of the given buffer.
Complete step by step answer:
Now from the question
Given, concentration of ${(N{H_4})_2}S{O_4}$=0.01M
Concentration of $N{H_4}OH$=0.02M
And $(p{K_a}$ of $NH_4^ + $ =9.26$)$
We know that pH = pK$_a$ + log$\dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}$
And pOH=pK$_b$ + log$\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$
$p{K_a}(NH_4^ + ) = 9.26$
$p{K_b}(N{H_3}) = 14 - 9.26 = 4.74$
$\left[ {N{H_4}} \right] = 2 \times \left[ {{{\left( {N{H_4}} \right)}_2}S{O_4}} \right]$
$ = 2 \times 0.001 = 0.02M$
$pOH = p{K_b} + \log \dfrac{{\left[ {NH_4^ + } \right]}}{{\left[ {N{H_4}OH} \right]}}$
$pOH = 4.74 + \log \dfrac{{0.02}}{{0.02}} = 4.74 - \log 1$
$
= 4.74 \\
pH = 14 - pOH = 14 - 4.74 \\
= 9.26 \\
$
Hence the pH of the given buffer is 9.26
So, the correct answer is “Option A”.
Note: In aqueous solution, an acid is defined as any species that increases the concentration of ${H^ + }$ aq and a base increases the concentration of $O{H^ - }$ aq. The concentrations of these ions in the solution can be very small. To avoid the very small numbers, these concentrations are converted into pH and pOH. Let's look at the definitions of pH and pOH
The pH for an aqueous solution is calculated from ${H^ + }$ using the following equation:
pH = - log $\left[ {{H^ + }} \right]$
The pOH for an aqueous solution is calculated from $O{H^ - }$ using the following equation:
pOH = - log $\left[ {O{H^ - }} \right]$
relation between pH and pOH is
pH + pOH = 14
A buffer solution is a solution which resists the change in it’s pH when an acid or base is added to it.
Acid buffer has acidic pH and is prepared by adding a weak acid and its salt with a strong base. Basic buffer has a basic pH and is prepared by adding a weak base and its salt with strong acid.
Complete step by step answer:
Now from the question
Given, concentration of ${(N{H_4})_2}S{O_4}$=0.01M
Concentration of $N{H_4}OH$=0.02M
And $(p{K_a}$ of $NH_4^ + $ =9.26$)$
We know that pH = pK$_a$ + log$\dfrac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}$
And pOH=pK$_b$ + log$\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$
$p{K_a}(NH_4^ + ) = 9.26$
$p{K_b}(N{H_3}) = 14 - 9.26 = 4.74$
$\left[ {N{H_4}} \right] = 2 \times \left[ {{{\left( {N{H_4}} \right)}_2}S{O_4}} \right]$
$ = 2 \times 0.001 = 0.02M$
$pOH = p{K_b} + \log \dfrac{{\left[ {NH_4^ + } \right]}}{{\left[ {N{H_4}OH} \right]}}$
$pOH = 4.74 + \log \dfrac{{0.02}}{{0.02}} = 4.74 - \log 1$
$
= 4.74 \\
pH = 14 - pOH = 14 - 4.74 \\
= 9.26 \\
$
Hence the pH of the given buffer is 9.26
So, the correct answer is “Option A”.
Note: In aqueous solution, an acid is defined as any species that increases the concentration of ${H^ + }$ aq and a base increases the concentration of $O{H^ - }$ aq. The concentrations of these ions in the solution can be very small. To avoid the very small numbers, these concentrations are converted into pH and pOH. Let's look at the definitions of pH and pOH
The pH for an aqueous solution is calculated from ${H^ + }$ using the following equation:
pH = - log $\left[ {{H^ + }} \right]$
The pOH for an aqueous solution is calculated from $O{H^ - }$ using the following equation:
pOH = - log $\left[ {O{H^ - }} \right]$
relation between pH and pOH is
pH + pOH = 14
A buffer solution is a solution which resists the change in it’s pH when an acid or base is added to it.
Acid buffer has acidic pH and is prepared by adding a weak acid and its salt with a strong base. Basic buffer has a basic pH and is prepared by adding a weak base and its salt with strong acid.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers