I have no idea what to do here....
What is the pH for the solution made from 11.10 mL of 0.2780 M NaOH solution mixed with 70.20 mL of 0.1300 M ammonium chloride solution? Note that the pKa for NH4+ is 9.241
To find the pH of the solution, we need to determine the concentration of the products that contribute to acidity or alkalinity. In this case, we have a mixture of a strong base (NaOH) and a weak acid (NH4+).
To start, let's calculate the moles of NaOH and NH4Cl using their molar concentrations and volumes:
Moles of NaOH = concentration × volume
= 0.2780 M × 11.10 mL
= 3.0828 mmol
Moles of NH4Cl = concentration × volume
= 0.1300 M × 70.20 mL
= 9.126 mmol
Next, we need to determine which compound is in excess. Since NH4Cl has more moles than NaOH, it will be the limiting reagent in this reaction.
NH4Cl will undergo hydrolysis, which produces NH3 (ammonia) and HCl (hydrochloric acid). The reaction can be represented as follows:
NH4+ + H2O ⇌ NH3 + H3O+
Since the pKa of NH4+ is given as 9.241, we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log([A-]/[HA])
In this case, [A-] is the concentration of NH3 and [HA] is the concentration of NH4+.
First, let's find the concentration of NH3:
As NH4+ reacts with water, it forms an equal amount of NH3 and H3O+. Therefore, the concentration of NH3 is equal to the moles of NH4Cl, which is 9.126 mmol, divided by the total volume of the solution, which is the sum of the volumes of NaOH and NH4Cl solutions:
Concentration of NH3 = moles of NH3 / total volume
= 9.126 mmol / (11.10 mL + 70.20 mL)
= (9.126 × 10^-3) / (81.30 × 10^-3) M
≈ 0.1120 M
Next, let's find the concentration of NH4+:
Since NH4+ is provided by NH4Cl and it dissociates into one NH4+ ion, the concentration of NH4+ is equal to the concentration of NH4Cl:
Concentration of NH4+ = concentration of NH4Cl
= 0.1300 M
Now, we can plug these values into the Henderson-Hasselbalch equation to find the pH:
pH = 9.241 + log([A-]/[HA])
= 9.241 + log(0.1120 M / 0.1300 M)
= 9.241 + log(0.8615)
≈ 9.241 + (-0.0644)
≈ 9.176
Therefore, the pH of the solution made from mixing 11.10 mL of 0.2780 M NaOH with 70.20 mL of 0.1300 M ammonium chloride is approximately 9.176.