Question

(a) Calculate the \(\mathrm{pH}\) of \(0.55 \mathrm{MHCN}\), if \(K_{\mathrm{a}}=6.2 \times 10^{-10}\) (b) Calculate the \(\mathrm{pOH}\) of \(0.044 \mathrm{M} \mathrm{HIO}_{3},\) if \(\bar{K}_{\mathrm{a}}=0.16\).

Short answer

The pH of 0.55 M HCN is approximately 5.24, and the pOH of 0.044 M HIO3 is approximately 12.92.

Step-by-step solution

Understand the problem

We need to calculate the pH of a weak acid solution HCN and the pOH of a weak acid solution HIO3 given their molarity and dissociation constants (Ka).

Write the expression for pH for HCN

For weak acids, the expression for the dissociation constant is \[K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\]. Here, HA is HCN, and A- is CN-. Since initial concentration of HCN is 0.55 M, and assuming \[ [\text{H}^+] = x, [\text{HCN}] = 0.55 - x\] and \[ [\text{CN}^-] = x\] where x is small:

Simplify for small x for HCN

Since \(x \ll 0.55\), it simplifies to: \[ K_a = \frac{x^2}{0.55}\]. Solve for x: \[ x^2 = K_a \times 0.55\]. Substitute the values: \[ x^2 = 6.2 \times 10^{-10} \times 0.55\].

Solve for x

Calculate x to find the concentration of H+: \[ x = \sqrt{6.2 \times 10^{-10} \times 0.55}\] gives \( x \approx 5.79 \times 10^{-6} \). The concentration of H+ is thus 5.79 \times 10^{-6} M.

Calculate pH for HCN

pH is given by \[ \text{pH} = -\text{log}[\text{H}^+]\]: \[ \text{pH} = -\text{log}(5.79 \times 10^{-6})\] gives \( \text{pH} \approx 5.24 \).

pH calculation for HIO3

For HIO3, write the expression for pH: \[ K_a = \frac{[\text{H}^+][\text{IO}_3^-]}{[\text{HIO}_3]}\]. Given initial concentration is 0.044 M. Assuming \[ [\text{H}^+] = x, [\text{HIO}_3] = 0.044 - x and [\text{IO}_3^-] = x \] simplifies to \[ K_a = \frac{x^2}{0.044}\]. Solve for x: \[ x^2 = 0.16 \times 0.044\].

Solve for x for HIO3

Calculate x to find the concentration of H+: \[ x = \sqrt{0.16 \times 0.044}\] gives \( x \approx 0.084 \). The concentration of H+ is thus 0.084 M.

Calculate pH and pOH for HIO3

pH is given by \[ \text{pH} = -\text{log}[\text{H}^+]\]: \[ \text{pH} = -\text{log}(0.084)\] gives \( \text{pH} \approx 1.08 \). Knowing \text{pH} + \text{pOH} = 14: \[ \text{pOH} = 14 - \text{pH}\]. Thus, \[ \text{pOH} = 14 - 1.08\], which simplifies to \[ \text{pOH} \approx 12.92\].

Key concepts

Weak Acid Dissociation

Weak acids only partially dissociate in water. This means not all the acid molecules ionize to release hydrogen ions (\text{H}^+) into the solution. When a weak acid, like HCN or HIO\text{3}, dissolves in water, it reaches an equilibrium between its non-dissociated form (HA) and its dissociated form (H\text{+} and A\text{-}).

This partial dissociation is described by the equilibrium equation:
\[ HA \rightleftharpoons H^+ + A^- \]
Where:

• HA is the weak acid.
• H\text{+} is the hydrogen ion.
• A\text{-} is the conjugate base of the acid.

Due to partial dissociation, only a small fraction of the weak acid molecules produce H\text{+} ions, making the concentration of H\text{+} relatively low in a weak acid solution.

Dissociation Constant (Ka)

The dissociation constant, \text{Ka}, measures the strength of a weak acid. It tells us how much the acid dissociates into H\text{+} and A\text{-} ions in water. The expression for \text{Ka} is given by:
\[ K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Here,

• [\text{H}\text{+}] is the concentration of hydrogen ions.
• [\text{A}\text{-}] is the concentration of the conjugate base.
• [\text{HA}] is the concentration of the non-dissociated acid.

Understanding \text{Ka} is crucial for solving pH problems involving weak acids because it allows us to predict the equilibrium concentrations of the ions in the solution. For instance, for HCN with \text{Ka} = 6.2 \times 10^{-10}, a small \text{Ka} value signifies very limited dissociation into H\text{+} and CN\text{-} ions.

Acid-Base Equilibrium

In acid-base chemistry, equilibrium refers to the state where the concentrations of reactants and products remain constant over time. For weak acids, this involves a balance between the non-dissociated acid (HA) and its ions (H\text{+} and A\text{-}). The equilibrium can be represented by the expression:
\[ HA \rightleftharpoons H^+ + A^- \]
To find equilibrium concentrations, we use the dissociation constant (\text{Ka}) and solve for unknowns such as H\text{+}. In our problem, we start by setting the initial concentration and assume a small degree of dissociation (x). We simplify assumptions (small x) and solve for the concentration of H\text{+}.

• For HCN: Initial concentration is 0.55 M. Small x simplifies to \text{Ka} = \frac{x^2}{0.55}. Solving for x gives concentration of H\text{+}.
• For HIO3: Initial concentration is 0.044 M. Small x simplifies to \text{Ka} = \frac{x^2}{0.044}. Solving for x gives concentration of H\text{+}.

pH and pOH relationship

The pH scale measures the acidity of a solution, with lower values indicating higher acidity (more H\text{+} ions). pH is calculated as:
\[ \text{pH} = -\text{log}[\text{H}^+] \]
For pOH, it measures the basicity (concentration of OH\text{-} ions), calculated as:
\[ \text{pOH} = -\text{log}[\text{OH}^-] \]
Importantly, pH and pOH are related through the equation:
\[ \text{pH} + \text{pOH} = 14 \]
This relationship helps solve for either pH or pOH if one knows the other. For instance, if the pH of HIO\text{3} is calculated to be 1.08, then:
\[ \text{pOH} = 14 - 1.08 = 12.92 \]
Thus, knowing the pH allows us to find the pOH and vice versa.