Question

A buffer solution is prepared in which the concentration of \(\mathrm{NH}_{3}\) is \(0.30 \mathrm{M}\) and the concentration of \(\mathrm{NH}_{4}{ }^{*}\) is \(0.20 \mathrm{M}\). If the equilibrium constant, \(\mathrm{K}_{\mathrm{b}}\) for \(\mathrm{NH}_{3}\) equals \(1.8 \times 10^{-5}\), what is the \(\mathrm{pH}\) of this solution ? ( \(\log 2.7=0.433\) ). (1) \(9.08\) (2) \(9.43\) (3) \(11.72\) (4) \(8.73\)

Short answer

The \(\mathrm{pH}\) of the solution is 9.43.

Step-by-step solution

Identify the given values

Given values: Concentration of \(\mathrm{NH}_{3}\): 0.30 M Concentration of \(\mathrm{NH}_{4}^{+}\): 0.20 M Equilibrium constant (\(\mathrm{K}_{b}\)): \(1.8 \times 10^{-5}\) \(\log 2.7 = 0.433\)

Write the expression for pOH using the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation for a base is: \[\mathrm{pOH} = \mathrm{p}K_{b} + \log \left( \frac{\left[\mathrm{NH}_4^{+}\right]}{\left[\mathrm{NH}_3\right]} \right)\]Where \(\mathrm{p}K_{b}\) is calculated from \(\mathrm{K}_{b}\) using \(\mathrm{p}K_{b} = -\log(\mathrm{K}_{b})\)

Calculate \(\mathrm{p}K_{b}\)

Calculate \(\mathrm{p}K_{b}\) using the given \(\mathrm{K}_{b}\): \[\mathrm{p}K_{b} = - \log (1.8 \times 10^{-5})\]Using properties of logarithms: \[\mathrm{p}K_{b} = - \left(\log 1.8 + \log 10^{-5}\right)\]\[\mathrm{p}K_{b} = - \left(0.255 + (-5)\right)\]\[\mathrm{p}K_{b} = 4.745\]

Calculate the ratio \(\dfrac{\left[\mathrm{NH}_{4}^{+}\right]}{\left[\mathrm{NH}_{3}\right]}\)

Substitute the given concentrations: \[\dfrac{\left[\mathrm{NH}_{4}^{+}\right]}{\left[\mathrm{NH}_{3}\right]} = \dfrac{0.20}{0.30} = 0.667\]

Calculate the pOH of the solution

Substitute \(\mathrm{p}K_{b}\) and the ratio into the Henderson-Hasselbalch equation: \[\mathrm{pOH} = 4.745 + \log (0.667) \]Given \(\log(2.7) = 0.433\)So, \(\log(\dfrac{1}{2.7}) = -0.433\) (since \(\log(\dfrac{1}{a}) = -\log(a)\)) \[\mathrm{pOH} = 4.745 + (-0.433)\]\[\mathrm{pOH} = 4.312\]

Calculate the pH of the solution

Use the relationship between pH and pOH: \[\mathrm{pH} + \mathrm{pOH} = 14\]\[\mathrm{pH} = 14 - \mathrm{pOH}\]\[\mathrm{pH} = 14 - 4.312 = 9.688\]The closest option to this calculated pH is: (2) 9.43

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a valuable formula in understanding buffer systems. It allows you to calculate the pH (or pOH) of a buffer solution based on the concentrations of its components. For a basic buffer, like the one involving ammonia (\text{NH}_3) and ammonium ion (\text{NH}_4^+), the equation is: \ \( \text{pOH} = \text{p}K_b + \text{log} \frac{[\text{NH}_4^+]}{[\text{NH}_3]} \)
The equation includes \text{p}K_b, which is the negative logarithm of the base dissociation constant (\text{K}_b). Here's why this is useful:

• It simplifies complex equilibrium expressions.
• By using it, you can quickly find pH if you know the concentrations of species in your buffer.

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Using given values, you can calculate the pOH of a solution and subsequently the pH, which is crucial for understanding the acidity/basicity of the solution.

Ammonia equilibrium

Ammonia (\text{NH}_3) acts as a weak base in water and establishes an equilibrium with ammonium ion (\text{NH}_4^+). This can be written as: \[ \text{NH}_3 + H_2O \rightleftharpoons \text{NH}_4^+ + OH^- \]
The equilibrium constant for this reaction is the base dissociation constant, \text{K}_b. For ammonia, given \text{K}_b is 1.8 \times 10^{-5}, which indicates ammonia's tendency to accept a proton and form \text{NH}_4^+. To solve pH problems involving ammonia:

• Identify the concentrations of \text{NH}_3 and \text{NH}_4^+ available.
• Calculate the ratio \( \frac{[\text{NH}_4^+]}{[\text{NH}_3]} \). In our exercise, it was 0.20 M / 0.30 M = 0.667.
• Insert these values into the Henderson-Hasselbalch equation to determine pOH.

Understanding ammonia's equilibrium helps in understanding how ammonia solutions resist changes in pH, thus proving their buffer properties.

pOH and pH relationship

pOH and pH are interconnected measures of a solution's acidity and basicity. The relationship is defined by the equation: \[ \text{pH} + \text{pOH} = 14 \]
This is due to the self-ionization of water which maintains a constant concentration product of hydrogen (\text{H}^+) and hydroxide (\text{OH}^-) ions. Knowing one measure can help calculate the other:

• If you know pOH, subtract it from 14 to get pH.
• This relationship is vital when you use the Henderson-Hasselbalch equation to find pOH first and then translate it to pH, as shown in our step-by-step solution.

For instance, once you have calculated pOH to be 4.312, you can easily find pH: \( \text{pH} = 14 - 4.312 = 9.688 \). This method ensures accurate and easy determination of the pH of the solution.