Question

A buffer is prepared by adding $7.00\mathrm{g}$ of ammonia $\left({\mathrm{NH}}_3\right)$ and $20.0\mathrm{g}$ of ammonium chloride $\left({\mathrm{NH}}_4\mathrm{Cl}\right)$ to enough water to form $2.50\mathrm{L}$ of solution. (a) What is the $\mathrm{pH}$ 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 9.28. (b) The complete ionic equation for the reaction with nitric acid is: NH${}_3$ (aq) + HNO${}_3$ (aq) → NH${}_4^{+}$ (aq) + NO${}_3^{-}$ (aq). (c) The complete ionic equation for the reaction with potassium hydroxide is: NH${}_4^{+}$ (aq) + KOH (aq) → NH${}_3$ (aq) + K^+ (aq) + H${}_2$O (l).

Step-by-step solution

Find the concentrations of NH${}_3$ and NH${}_4$Cl in the solution

To find the pH of the buffer, we need the concentrations of NH${}_3$ and NH${}_4$Cl in the solution. First, we'll use the molar masses of NH${}_3$ and NH${}_4$Cl to determine the moles of both substances in the solution. Then, we'll divide the moles by the volume of the solution (2.50 L) to find the concentrations. Molar mass of NH${}_3$ = 17.03 g/mol Moles of NH${}_3$ = (7.00 g) / (17.03 g/mol) = 0.411 mol Molar mass of NH${}_4$Cl = 53.49 g/mol Moles of NH${}_4$Cl = (20.0 g) / (53.49 g/mol) = 0.374 mol Concentration of NH${}_3$ = (0.411 mol) / (2.50 L) = 0.164 M Concentration of NH${}_4$Cl = (0.374 mol) / (2.50 L) = 0.150 M

Calculate pH using the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH to the pK_a, and the concentrations of the weak acid and its conjugate base: pH = pK_a + log ([A-] / [HA]) In this case, [HA] refers to the concentration of NH${}_4^{+}$ and [A-] refers to the concentration of NH${}_3$. The pK_a value for NH${}_4^{+}$ is 9.25. Now, we can solve for the pH of the buffer solution: pH = 9.25 + log (0.164 / 0.150) = 9.25 + 0.038 = 9.28 So, the pH of the buffer solution is 9.28.

Write the complete ionic equation for nitric acid reaction

When nitric acid (HNO${}_3$) is added to the buffer, it reacts with NH${}_3$ to form NH${}_4^{+}$ and NO${}_3^{-}$ ions: NH${}_3$ (aq) + HNO${}_3$ (aq) -> NH${}_4^{+}$ (aq) + NO${}_3^{-}$ (aq) The complete ionic equation is the same as the molecular equation in this case because all ions are already included.

Write the complete ionic equation for potassium hydroxide reaction

When potassium hydroxide (KOH) is added to the buffer, it reacts with NH${}_4^{+}$ to form NH${}_3$ and K^+ and OH^- ions: NH${}_4^{+}$ (aq) + OH^- (aq) -> NH${}_3$ (aq) + H${}_2$O (l) K^+ ions do not participate in the reaction, so they are spectator ions. Now, we can include the spectator ions in the complete ionic equation: NH${}_4^{+}$ (aq) + KOH (aq) -> NH${}_3$ (aq) + K^+ (aq) + H${}_2$O (l) In summary: (a) The pH of the buffer solution is 9.28. (b) The complete ionic equation for the reaction with nitric acid is: NH${}_3$ (aq) + HNO${}_3$ (aq) -> NH${}_4^{+}$ (aq) + NO${}_3^{-}$ (aq). (c) The complete ionic equation for the reaction with potassium hydroxide is: NH${}_4^{+}$ (aq) + KOH (aq) -> NH${}_3$ (aq) + K^+ (aq) + H${}_2$O (l).