
The compound ${A_2}S{O_4}$ formed by $AOH$ and ${H_2}S{O_4}$. Calculate $pH$ when $p{K_b}$ of $AOH = 12$. Please describe it briefly.
Answer
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Hint: We can define pH as the power of hydrogen ions in a given solution. We can calculate pH of the solution using a formula,
$pH = - \log 10\left[ {{H^ + }} \right]$
If the pH of the solution is 0 then the solution is highly acidic, 14 means the solution is highly basic. The pH value of 7 shows as neutral as water.
Complete answer:
Let us discuss the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation gives a connection among the \[pH\] of acids and their \[pKa\] (acid dissociation constant). The \[pH\] of a buffer solution can be predicted with the help of this equation if the concentration of the acid and its conjugate base, or the base and the corresponding conjugate acid, are known. The Henderson-Hasselbalch equation can be given as,
\[pOH = pKb + log10\left( {\dfrac{{\left[ {Acid} \right]}}{{\left[ {Base} \right]}}} \right)\]
The given value of ${\text{p}}{{\text{K}}_{\text{b}}}{\text{ = 12}}$
The dissociation equation is,
${H_2}S{O_4} + 2AOH \to {A_2}S{O_4} + 2{H_2}O$
From the equation we are able to know that if we assume the concentration of sulfuric acid as $0.1\,M$ then the concentration of $AOH$ is half of the concentration of sulfuric acid.
Now, we calculate the $pOH$ of the solution as below,
$pOH = 12 + \log \left( {\dfrac{{0.1}}{{0.05}}} \right)$
$ \Rightarrow pOH = 12 + \log (2)$
Substituting the log value we get,
$ \Rightarrow pOH = 12 + 0.301$
On adding these values we get,
$ \Rightarrow pOH = 12.301$
Now we can able to calculate the $pH$ value using the relation,
$pOH + pH = 14$
$pH = 14 - pOH$
On substituting the value of pOH we get,
$pH = 14 - 12.301 = 1.69$
The $pH$ of the solution is $1.69$.
Note:
-We define indicators as weak acids that exist as natural dyes and indicate the concentration of ions during a solution via color change. The value of $pH$ is determined from the negative logarithm of this concentration and is engaged to point to the acidic, basic, or neutral nature of the substance you're testing.
$pH = - \log 10\left[ {{H^ + }} \right]$
If the pH of the solution is 0 then the solution is highly acidic, 14 means the solution is highly basic. The pH value of 7 shows as neutral as water.
Complete answer:
Let us discuss the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation gives a connection among the \[pH\] of acids and their \[pKa\] (acid dissociation constant). The \[pH\] of a buffer solution can be predicted with the help of this equation if the concentration of the acid and its conjugate base, or the base and the corresponding conjugate acid, are known. The Henderson-Hasselbalch equation can be given as,
\[pOH = pKb + log10\left( {\dfrac{{\left[ {Acid} \right]}}{{\left[ {Base} \right]}}} \right)\]
The given value of ${\text{p}}{{\text{K}}_{\text{b}}}{\text{ = 12}}$
The dissociation equation is,
${H_2}S{O_4} + 2AOH \to {A_2}S{O_4} + 2{H_2}O$
From the equation we are able to know that if we assume the concentration of sulfuric acid as $0.1\,M$ then the concentration of $AOH$ is half of the concentration of sulfuric acid.
Now, we calculate the $pOH$ of the solution as below,
$pOH = 12 + \log \left( {\dfrac{{0.1}}{{0.05}}} \right)$
$ \Rightarrow pOH = 12 + \log (2)$
Substituting the log value we get,
$ \Rightarrow pOH = 12 + 0.301$
On adding these values we get,
$ \Rightarrow pOH = 12.301$
Now we can able to calculate the $pH$ value using the relation,
$pOH + pH = 14$
$pH = 14 - pOH$
On substituting the value of pOH we get,
$pH = 14 - 12.301 = 1.69$
The $pH$ of the solution is $1.69$.
Note:
-We define indicators as weak acids that exist as natural dyes and indicate the concentration of ions during a solution via color change. The value of $pH$ is determined from the negative logarithm of this concentration and is engaged to point to the acidic, basic, or neutral nature of the substance you're testing.
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