What is the pH of the solution after 50.0 mL of NaOH was added to a 200.0 mL solution of 0.40 M ammonium chloride?
Understand the Problem
The question is asking us to calculate the pH of a solution after a specified volume of sodium hydroxide (NaOH) has been added to a solution of ammonium chloride. This calculation typically requires using the concepts of weak acids/alkalis and determining the concentration of hydroxide ions after reaction to find the final pH.
Answer
The pH can be determined by following the steps above, and requires specific volumes and concentrations for a final numeric answer.
Answer for screen readers
The pH of the solution after the specified volumes of NaOH and ammonium chloride are mixed can be calculated using the formulas above, depending on the specific values of volumes and concentrations given.
Steps to Solve
- Identify Initial Information
Begin by noting the concentrations and the volumes of the solutions involved. For example, let ( V_{NH4Cl} ) be the volume of ammonium chloride solution, ( C_{NH4Cl} ) its concentration, ( V_{NaOH} ) the volume of sodium hydroxide, and ( C_{NaOH} ) its concentration.
- Calculate Moles of Ammonium Chloride
Using the initial concentration and volume of ammonium chloride: $$ \text{Moles of } NH_4Cl = C_{NH4Cl} \times V_{NH4Cl} $$
- Calculate Moles of Sodium Hydroxide
Calculate the moles of sodium hydroxide using its concentration and volume: $$ \text{Moles of } NaOH = C_{NaOH} \times V_{NaOH} $$
- Determine Reaction Between NaOH and NH4Cl
The reaction that takes place is: $$ NH_4^+ + OH^- \rightleftharpoons NH_3 + H_2O $$
Because NaOH is a strong base, it will react with the weak acid ( NH_4^+ ).
- Calculate Remaining Moles After Reaction
Subtract the moles of ( NaOH ) from ( NH_4^+ ).
- If ( \text{Moles of } NaOH ) < ( \text{Moles of } NH_4Cl ):
- Remaining ( NH_4^+ = \text{Moles of } NH_4Cl - \text{Moles of } NaOH )
- ( OH^- ) ions will be produced equal to the moles of ( NaOH ).
- If ( \text{Moles of } NaOH ) > ( \text{Moles of } NH_4Cl ):
- Remaining ( OH^- = \text{Moles of } NaOH - \text{Moles of } NH_4Cl )
- Calculate Final Concentration of Ions
Calculate final concentrations:
- For ( NH_4^+ ), ( [NH_4^+] = \frac{\text{Remaining Moles of } NH_4^+}{\text{Total Volume}} )
- For ( OH^- ), ( [OH^-] = \frac{\text{Remaining Moles of } OH^-}{\text{Total Volume}} )
- Calculate pOH and pH
To find pOH: $$ pOH = -\log[OH^-] $$
And then use it to find pH: $$ pH = 14 - pOH $$
The pH of the solution after the specified volumes of NaOH and ammonium chloride are mixed can be calculated using the formulas above, depending on the specific values of volumes and concentrations given.
More Information
The pH of the solution reflects the balance between the weak acid (ammonium ion) and strong base (sodium hydroxide). As you alter the amounts of these substances, the pH will change, showcasing the principles of acid-base neutralization.
Tips
- Forgetting to convert volumes to liters before calculating moles.
- Confusing the roles of strong and weak acids/bases in the reaction.
- Not considering dilution when calculating final concentrations of ions after mixing.
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