The \[pKa\] of a weak acid \[HA\] is \[4.5\]. The \[pOH\] of an aqueous buffered solution of \[50\% \] ionized \[HA\] will be:
A.\[4.5\]
B.\[2.5\]
C.\[9.5\]
D.\[7.0\]
Answer
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Hint: To answer this question, you should recall the concept of the Henderson-Hasselbalch Equation. Use the values in the question given and apply it in the equation.
Formula used:
Using, Henderson-Hasselbalch Equation: \[ \to {\text{ }}pH = p{K_a} + \log \dfrac{{\left[ {{\text{conjugate base}}} \right]}}{{\left[ {{\text{acid}}} \right]}}\]
Complete step by step answer:
- We know that equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. The Henderson-Hasselbalch equation provides a relationship between the pH of acids (in aqueous solutions) and their \[{\text{pKa}}\] (acid dissociation constant). The pH of a solution is often estimated with the assistance of this equation when the concentration of the acid and its conjugate base, or the bottom and the corresponding conjugate acid, are known.
- The pH of Buffer Solutions shows minimal change upon the addition of a very small quantity of strong acid or strong base. They are therefore used to keep the pH at a constant value.
- Using the values given in the question and using in \[pH = pKa + log\dfrac{{\left[ {{\text{Conjugate base}}} \right]}}{{\left[ {{\text{acid}}} \right]}}\]
We know the value of \[p{K_a} = 4.5\] and \[\left[ {{\text{Conjugate base}}} \right]{\text{ = }}\left[ {{\text{acid}}} \right]{\text{,}}\] as \[50\% \] ionisation.
\[\therefore pH = 4.5 + log\dfrac{{0.5}}{{0.5}}\].
Hence we will get \[pH = 4.5\].
We know that \[\left( {pH + pOH = 14} \right)\].
Therefore the value of \[pOH\] can be calculated as:
\[ \Rightarrow \;pOH = 14 - 4.5 = 9.5\]
Thus, the correct option is C.
Note:
- Henderson-Hasselbalch equation is used to find the pH of buffers. You should know that the Henderson-Hasselbalch equation fails to predict accurate values for the strong acids and strong bases because it assumes that the concentration of the acid and its conjugate base at equilibrium will remain an equivalent because of the formal concentration.
- Since the Henderson-Hasselbalch equation doesn't consider the self-dissociation undergone by water, it fails to supply accurate pH values for very dilute buffer solutions. You should remember the difference between strong bases and weak bases: a strong base is a base that is ionized in solution but if it is less than ionized in solution, it is a weak base.
- There are only a few strong bases and certain salts also will affect the acidity or basicity of aqueous solutions because a number of the ions will undergo hydrolysis. The general rule is that salts with ions that are part of strong acids or bases will not hydrolyse, while salts with ions that are part of weak acids or bases will hydrolyse.
Formula used:
Using, Henderson-Hasselbalch Equation: \[ \to {\text{ }}pH = p{K_a} + \log \dfrac{{\left[ {{\text{conjugate base}}} \right]}}{{\left[ {{\text{acid}}} \right]}}\]
Complete step by step answer:
- We know that equilibrium constant for any reaction is calculated by the ratio of the concentration of products to that of reactant. The Henderson-Hasselbalch equation provides a relationship between the pH of acids (in aqueous solutions) and their \[{\text{pKa}}\] (acid dissociation constant). The pH of a solution is often estimated with the assistance of this equation when the concentration of the acid and its conjugate base, or the bottom and the corresponding conjugate acid, are known.
- The pH of Buffer Solutions shows minimal change upon the addition of a very small quantity of strong acid or strong base. They are therefore used to keep the pH at a constant value.
- Using the values given in the question and using in \[pH = pKa + log\dfrac{{\left[ {{\text{Conjugate base}}} \right]}}{{\left[ {{\text{acid}}} \right]}}\]
We know the value of \[p{K_a} = 4.5\] and \[\left[ {{\text{Conjugate base}}} \right]{\text{ = }}\left[ {{\text{acid}}} \right]{\text{,}}\] as \[50\% \] ionisation.
\[\therefore pH = 4.5 + log\dfrac{{0.5}}{{0.5}}\].
Hence we will get \[pH = 4.5\].
We know that \[\left( {pH + pOH = 14} \right)\].
Therefore the value of \[pOH\] can be calculated as:
\[ \Rightarrow \;pOH = 14 - 4.5 = 9.5\]
Thus, the correct option is C.
Note:
- Henderson-Hasselbalch equation is used to find the pH of buffers. You should know that the Henderson-Hasselbalch equation fails to predict accurate values for the strong acids and strong bases because it assumes that the concentration of the acid and its conjugate base at equilibrium will remain an equivalent because of the formal concentration.
- Since the Henderson-Hasselbalch equation doesn't consider the self-dissociation undergone by water, it fails to supply accurate pH values for very dilute buffer solutions. You should remember the difference between strong bases and weak bases: a strong base is a base that is ionized in solution but if it is less than ionized in solution, it is a weak base.
- There are only a few strong bases and certain salts also will affect the acidity or basicity of aqueous solutions because a number of the ions will undergo hydrolysis. The general rule is that salts with ions that are part of strong acids or bases will not hydrolyse, while salts with ions that are part of weak acids or bases will hydrolyse.
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