Answer
Verified
388.5k+ views
Hint: pH of a solution is calculated from the concentration of ${{H}^{+}}$ ions in the solution and is given as $pH=-\log \left[ {{H}^{+}} \right]$.
Similarly, pOH measures $O{{H}^{-}}$ ion concentration in a solution, i.e. $pOH=-\log \left[ O{{H}^{-}} \right]$.
Relationship between pH and pOH, based on the equilibrium concentrations of ${{H}^{+}}$ and $O{{H}^{-}}$ is
\[pH+pOH=14\]
Complete answer:
Let us solve the given question step by step.
Given, molar concentration of NaOH base = ${{10}^{-6}}$ M
NaOH being a strong base dissociates completely in water into $N{{a}^{+}}$ and $O{{H}^{-}}$ ions
$NaOH(aq)\to N{{a}^{+}}(aq)+O{{H}^{-}}(aq)$
Since, one mole of NaOH is equal to one of $N{{a}^{+}}$ and $O{{H}^{-}}$. Thus, we can draw the following conclusion
\[\left[ NaOH \right]=\left[ N{{a}^{+}} \right]=\left[ O{{H}^{-}} \right]={{10}^{-6}}M\]
It is given that the base has been diluted by 100 times. So, the concentration of $\left[ O{{H}^{-}} \right]$now becomes ${{10}^{-8}}M$, i.e.
\[\dfrac{{{10}^{-6}}M}{100}={{10}^{-8}}M\]
To find the total concentration of $O{{H}^{-}}$ ions, i.e. $\left[ O{{H}^{-}} \right]={{10}^{-8}}M+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}$, we need to consider the concentration of $O{{H}^{-}}$ from water.
We know that water ionizes as
\[{{H}_{2}}O\rightleftarrows {{H}^{+}}+O{{H}^{-}}\]
One mole of water gives one mole of ${{H}^{+}}$ and $O{{H}^{-}}$,thus, we have
\[\left[ {{H}^{+}} \right]=\left[ O{{H}^{-}} \right]\]
Ionic product of water, which is the product of concentration of ${{H}^{+}}$ and $O{{H}^{-}}$ ions, at 25$^{o}C$ is ${{10}^{-14}}$. Since \[\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}\], we can write that the concentration of $O{{H}^{-}}$ from ionization of water, ${{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}={{10}^{-7}}M$.
Therefore, total $\left[ O{{H}^{-}} \right]$ ions after substituting the value of ${{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}$ will be
$\begin{align}
& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}} \\
& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}} \\
\end{align}$
Multiplying and diving ${{10}^{-7}}$ by 10 in the above equation, we get
$\begin{align}
& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}}\times \frac{10}{10} \\
& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-8}}\times 10 \\
\end{align}$
Taking ${{10}^{-8}}$ common in the equation for simplification, we obtain
$\begin{align}
& \left[ O{{H}^{-}} \right]={{10}^{-8}}(1+10) \\
& \left[ O{{H}^{-}} \right]=11\times {{10}^{-8}} \\
\end{align}$
Now we have the total concentration of $O{{H}^{-}}$, i.e. $\left[ O{{H}^{-}} \right]=11\times {{10}^{-8}}M$, we can find pOH as
$\begin{align}
& pOH=-\log \left[ O{{H}^{-}} \right] \\
& pOH=-\log \left[ 11\times {{10}^{-8}} \right] \\
\end{align}$
Applying $\log (mn)=\log m+\log n$ and $\log {{m}^{n}}=n\log m$, we can simplify the above equation as
\[\begin{align}
& pOH=-(\log 11+\log {{10}^{-8}}) \\
& pOH=-(\log 11-8\log 10) \\
\end{align}\]
We know that ${{\log }_{10}}10=1$, on substituting it, the above equation becomes
\[\begin{align}
& pOH=-(\log 11-8) \\
& pOH=-1.0414+8 \\
& pOH=6.9586\approx 6.96 \\
\end{align}\]
To find the pH from pOH, we have the relation that is true for solution at 25$^{o}C$. Putting the value of pOH = 6.96, we have the pH of the solution
\[\begin{align}
& pH+pOH=14 \\
& pH=14-pOH \\
& pH=14-6.96 \\
& pH=7.04 \\
\end{align}\]
Therefore, the pH of diluted base is 7.04, which lies within 7 to 8.
So, the correct answer is “Option C”.
Note: Note that concentration of ${{H}^{+}}$ and $O{{H}^{-}}$ is important for dilute solutions. We cannot ignore the concentration of $O{{H}^{-}}$ due to water in this case, as the solution has become very dilute due to the addition of water. Due to the decrease in the number of $O{{H}^{-}}$ (and ${{H}^{+}}$) ions per unit volume, the pH of the basic solution has been reduced.
Similarly, pOH measures $O{{H}^{-}}$ ion concentration in a solution, i.e. $pOH=-\log \left[ O{{H}^{-}} \right]$.
Relationship between pH and pOH, based on the equilibrium concentrations of ${{H}^{+}}$ and $O{{H}^{-}}$ is
\[pH+pOH=14\]
Complete answer:
Let us solve the given question step by step.
Given, molar concentration of NaOH base = ${{10}^{-6}}$ M
NaOH being a strong base dissociates completely in water into $N{{a}^{+}}$ and $O{{H}^{-}}$ ions
$NaOH(aq)\to N{{a}^{+}}(aq)+O{{H}^{-}}(aq)$
Since, one mole of NaOH is equal to one of $N{{a}^{+}}$ and $O{{H}^{-}}$. Thus, we can draw the following conclusion
\[\left[ NaOH \right]=\left[ N{{a}^{+}} \right]=\left[ O{{H}^{-}} \right]={{10}^{-6}}M\]
It is given that the base has been diluted by 100 times. So, the concentration of $\left[ O{{H}^{-}} \right]$now becomes ${{10}^{-8}}M$, i.e.
\[\dfrac{{{10}^{-6}}M}{100}={{10}^{-8}}M\]
To find the total concentration of $O{{H}^{-}}$ ions, i.e. $\left[ O{{H}^{-}} \right]={{10}^{-8}}M+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}$, we need to consider the concentration of $O{{H}^{-}}$ from water.
We know that water ionizes as
\[{{H}_{2}}O\rightleftarrows {{H}^{+}}+O{{H}^{-}}\]
One mole of water gives one mole of ${{H}^{+}}$ and $O{{H}^{-}}$,thus, we have
\[\left[ {{H}^{+}} \right]=\left[ O{{H}^{-}} \right]\]
Ionic product of water, which is the product of concentration of ${{H}^{+}}$ and $O{{H}^{-}}$ ions, at 25$^{o}C$ is ${{10}^{-14}}$. Since \[\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}\], we can write that the concentration of $O{{H}^{-}}$ from ionization of water, ${{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}={{10}^{-7}}M$.
Therefore, total $\left[ O{{H}^{-}} \right]$ ions after substituting the value of ${{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}}$ will be
$\begin{align}
& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{\left[ O{{H}^{-}} \right]}_{{{H}_{2}}O}} \\
& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}} \\
\end{align}$
Multiplying and diving ${{10}^{-7}}$ by 10 in the above equation, we get
$\begin{align}
& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-7}}\times \frac{10}{10} \\
& \left[ O{{H}^{-}} \right]={{10}^{-8}}+{{10}^{-8}}\times 10 \\
\end{align}$
Taking ${{10}^{-8}}$ common in the equation for simplification, we obtain
$\begin{align}
& \left[ O{{H}^{-}} \right]={{10}^{-8}}(1+10) \\
& \left[ O{{H}^{-}} \right]=11\times {{10}^{-8}} \\
\end{align}$
Now we have the total concentration of $O{{H}^{-}}$, i.e. $\left[ O{{H}^{-}} \right]=11\times {{10}^{-8}}M$, we can find pOH as
$\begin{align}
& pOH=-\log \left[ O{{H}^{-}} \right] \\
& pOH=-\log \left[ 11\times {{10}^{-8}} \right] \\
\end{align}$
Applying $\log (mn)=\log m+\log n$ and $\log {{m}^{n}}=n\log m$, we can simplify the above equation as
\[\begin{align}
& pOH=-(\log 11+\log {{10}^{-8}}) \\
& pOH=-(\log 11-8\log 10) \\
\end{align}\]
We know that ${{\log }_{10}}10=1$, on substituting it, the above equation becomes
\[\begin{align}
& pOH=-(\log 11-8) \\
& pOH=-1.0414+8 \\
& pOH=6.9586\approx 6.96 \\
\end{align}\]
To find the pH from pOH, we have the relation that is true for solution at 25$^{o}C$. Putting the value of pOH = 6.96, we have the pH of the solution
\[\begin{align}
& pH+pOH=14 \\
& pH=14-pOH \\
& pH=14-6.96 \\
& pH=7.04 \\
\end{align}\]
Therefore, the pH of diluted base is 7.04, which lies within 7 to 8.
So, the correct answer is “Option C”.
Note: Note that concentration of ${{H}^{+}}$ and $O{{H}^{-}}$ is important for dilute solutions. We cannot ignore the concentration of $O{{H}^{-}}$ due to water in this case, as the solution has become very dilute due to the addition of water. Due to the decrease in the number of $O{{H}^{-}}$ (and ${{H}^{+}}$) ions per unit volume, the pH of the basic solution has been reduced.
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
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE