Question

A buffer is prepared by adding 10.0 \(\mathrm{g}\) of ammonium chloride \(\left(\mathrm{NH}_{4} \mathrm{Cl}\right)\) to 250 \(\mathrm{mL}\) of 1.00 \(\mathrm{M} \mathrm{NH}_{3}\) solution. (a) What is the pH of this buffer? (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

Short answer

\(a)\) The pH of the buffer solution is 9.376. \(b)\) The complete ionic equation for the reaction with nitric acid is: NH3(aq) + H\(^{+}\)(aq) + NO3\(^{-}\)(aq) -> NH4\(^{+}\)(aq) + NO3\(^{-}\)(aq). \(c)\) The complete ionic equation for the reaction with potassium hydroxide is: NH4\(^{+}\)(aq) + OH\(^{-}\)(aq) -> NH3(aq) + H2O(l).

Step-by-step solution

Calculate the moles of NH4Cl and NH3

First, we need to find the moles of ammonium chloride and ammonia present in the buffer solution. Moles of NH4Cl = mass / molecular weight = 10.0 g / 53.5 g/mol = 0.18692 mol NH3 solution's volume = 250 mL = 0.250 L Moles of NH3 = concentration × volume = 1.00 M × 0.250 L = 0.250 mol

Apply the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: pH = pKa + log ([A-] / [HA]) In this case, the conjugate acid is NH4+ (ammonium ion) and the conjugate base is NH3 (ammonia). The Ka of NH4+ is 5.56 × 10^(-10), so the pKa is given by: pKa = -log(5.56 × 10^(-10)) = 9.25 Now we can substitute the values of the pKa, moles of NH3 (A-) and moles of NH4+ (HA) in the equation: pH = 9.25 + log((0.250 mol) / (0.18692 mol))

Calculate the pH

Using the given values, we can now calculate the pH of the buffer solution: pH = 9.25 + log(1.337) = 9.25 + 0.126 = 9.376 So the pH of the buffer solution is 9.376. #b) Complete ionic equation when nitric acid is added#

Write the balanced molecular equation

When nitric acid (HNO3) is added to the buffer solution containing NH3 (ammonia) and NH4+ (ammonium ion), it produces NH4+ (ammonium ion) and NO3- (nitrate ion). NH3 + HNO3 -> NH4+ + NO3-

Separate the ions and write the complete ionic equation

Separating the strong electrolytes into their respective ions, we get the ionic equation: NH3 + H+ + NO3- -> NH4+ + NO3- The complete ionic equation for this reaction is: NH3(aq) + H+(aq) + NO3-(aq) -> NH4+(aq) + NO3-(aq) #c) Complete ionic equation when potassium hydroxide is added#

Write the balanced molecular equation

When potassium hydroxide (KOH) is added to the buffer solution containing NH4+ (ammonium ion) and NH3 (ammonia), it produces K+ (potassium ion), NH3 (ammonia), and H2O (water). NH4+ + OH- -> NH3 + H2O

Separate the ions and write the complete ionic equation

Separating the strong electrolytes into their respective ions, we get the ionic equation: NH4+ + OH- -> NH3 + H2O The complete ionic equation for this reaction is: NH4+(aq) + OH-(aq) -> NH3(aq) + H2O(l)

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a fundamental formula used to calculate the pH of buffer solutions. It is expressed as:\[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right),\]where:

• \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\).
• \([A^-]\) is the concentration of the base form of the buffer.
• \([HA]\) is the concentration of the acid form of the buffer.

In a buffer solution composed of ammonia \((NH_3)\) and ammonium chloride \((NH_4Cl)\), ammonia serves as the base \((A^-)\), while ammonium \((NH_4^+)\) is the conjugate acid \((HA)\). The stability of the pH is maintained by the resistive properties of these components against the addition of strong acids or bases.
To determine buffer pH, calculate \(pK_a\) using the relation: \(pK_a = -\log(K_a)\). For example, if \(K_a\) of \(NH_4^+\) is \(5.56 \times 10^{-10}\), then \(pK_a = 9.25\).
Next, input the concentrations (or moles) into the equation, providing insight into how these values influence the pH levels, crucial in fields such as biochemistry and medicine where pH control is necessary for reactions.

pH calculation

Calculating pH, particularly for a buffer solution, involves using the concentrations of its acidic and basic components. For instance, in a buffer made from \(NH_3\) and \(NH_4Cl\), the solution involves finding the number of moles of both compounds. These steps include:

• Converting grams of \(NH_4Cl\) to moles: \(\text{moles of } NH_4Cl = \frac{10.0\, \text{g}}{53.5\, \text{g/mol}} = 0.18692\, \text{mol}\).
• Calculating moles of \(NH_3\) using its concentration: \(\text{moles of } NH_3 = 1.00\, \text{M} \times 0.250\, \text{L} = 0.250\, \text{mol}\).

Using these values in the Henderson-Hasselbalch equation provides the pH:\[pH = 9.25 + \log\left(\frac{0.250}{0.18692}\right)\]
Simplifying, \(\log(1.337) = 0.126\), hence:\[pH = 9.25 + 0.126 = 9.376\]This calculated pH illustrates how buffers moderate the effect of added acids or bases, keeping pH relatively constant.

Ionic equations

Ionic equations offer a perspective on how ions participate in chemical reactions, particularly in solutions. When acids or bases are added to a buffer system, an ionic equation illustrates the change at the ionic level:
When nitric acid \((HNO_3)\) is added to an ammonia buffer system:

• The nitric acid dissociates into \(H^+\) and \(NO_3^-\) ions.
• These ions react with ammonia \((NH_3)\) forming additional ammonium ions \((NH_4^+)\).

The complete ionic equation for the addition of \(HNO_3\) is:\[NH_3(aq) + H^+(aq) + NO_3^-(aq) \rightarrow NH_4^+(aq) + NO_3^-(aq)\]
Similarly, when potassium hydroxide \((KOH)\) is added:

• The \(OH^-\) ions react with \(NH_4^+\) to reform \(NH_3\) and water \((H_2O)\).

The ionic equation then is:\[NH_4^+(aq) + OH^-(aq) \rightarrow NH_3(aq) + H_2O(l)\]These equations highlight the role of buffers in maintaining pH stability by neutralizing added acids or bases through reactions with ions present in the solution.