
The pH of 0.01M ${(N{H_4})_2}S{O_4}$ and 0.02M $N{H_4}OH$ buffer solution $(p{K_a}$ of $NH_4^ + $ = 9.26$)$ is:
A. 9.26
B. 4.74
C. 4.74 + log2
D. None of these
Answer
586.8k+ 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.
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