KCET · Chemistry · Ionic Equilibrium
Which among the following has highest \(\mathrm{pH}\) ?
- A \(1 \mathrm{M} \mathrm{NaOH}\)
- B \(1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4\)
- C \(0.1 \mathrm{M} \mathrm{NaOH}\)
- D \(1 \mathrm{M} \mathrm{HCl}\)
Answer & Solution
Correct Answer
(A) \(1 \mathrm{M} \mathrm{NaOH}\)
Step-by-step Solution
Detailed explanation
For acidic medium, we know that \(\mathrm{pH}\) values range is between \(0-6.9\), for neutral it is 7 and for bases it is \(7.1-14\). So, \(\mathrm{HCl}\) is acidic and \(\mathrm{NaOH}\) is basic. Thus, we can straight forward infer that \(\mathrm{NaOH}\) has a higher \(\mathrm{pH}\) value.
Now, 1 mole of sodium hydroxide produces 1 mole of \(\mathrm{Na}^{+}\)and 1 mole of \(\mathrm{OH}^{-}\)ions. This means that the concentration of the hydroxide ions will be equal to \(\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=1 \mathrm{M}\)
For case \(\mathrm{A}\),
\[
\begin{aligned}
& \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log [1]=0 \\
& \therefore \quad \mathrm{pH}=14-\mathrm{pOH}=14-0=14 \\
& \text { For case } B \text {, } \\
& {\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=0.1 \mathrm{M}} \\
& \text { pOH }=-\log [0.1]=1 \\
& \therefore \quad \mathrm{pH}=14-1=13 \\
&
\end{aligned}
\]
So, pH of \(1 \mathrm{M} \mathrm{NaOH}\) is highest among the given solution.
Now, 1 mole of sodium hydroxide produces 1 mole of \(\mathrm{Na}^{+}\)and 1 mole of \(\mathrm{OH}^{-}\)ions. This means that the concentration of the hydroxide ions will be equal to \(\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=1 \mathrm{M}\)
For case \(\mathrm{A}\),
\[
\begin{aligned}
& \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log [1]=0 \\
& \therefore \quad \mathrm{pH}=14-\mathrm{pOH}=14-0=14 \\
& \text { For case } B \text {, } \\
& {\left[\mathrm{OH}^{-}\right]=[\mathrm{NaOH}]=0.1 \mathrm{M}} \\
& \text { pOH }=-\log [0.1]=1 \\
& \therefore \quad \mathrm{pH}=14-1=13 \\
&
\end{aligned}
\]
So, pH of \(1 \mathrm{M} \mathrm{NaOH}\) is highest among the given solution.
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