You are asked to bring the pH of 0.500L of 0.450M NH4Cl to 7.00.
a) Which of the following solutions would you use: 10.0M HCl or 10.0M NH3?
I figured that one would use the 10.0M NH3
b)How many drops (1 drop 0.05mL) of this solution would you use?
To determine whether you would use 10.0M HCl or 10.0M NH3 to bring the pH of the NH4Cl solution to 7.00, you need to understand the chemistry behind the problem.
a) NH4Cl is an acidic salt since it is the product of a weak base (NH3) reacting with a strong acid (HCl). NH4Cl dissociates in water to release NH4+ ions and Cl- ions. The NH4+ ions can undergo hydrolysis in water, leading to the formation of NH3 and H3O+ ions. Since NH3 is a weak base, it can react with the H3O+ ions and increase the pH of the solution.
To increase the pH to 7.00, you will need to add a base to neutralize the acidic H3O+ ions. Therefore, you would use 10.0M NH3, as it is a basic solution that can react with the H3O+ ions and raise the pH.
b) To determine how many drops of 10.0M NH3 solution you would need, you need to know the volume of 10.0M NH3 solution required to raise the pH to 7.00.
First, calculate the moles of NH4Cl in 0.500L of 0.450M NH4Cl using the equation:
moles = concentration (M) × volume (L)
moles of NH4Cl = 0.450M × 0.500L
moles of NH4Cl = 0.225 moles
Since NH4Cl is a 1:1 electrolyte, the moles of H3O+ ions are also 0.225 moles.
Next, calculate the moles of NH3 required to react with the H3O+ ions to neutralize them and raise the pH to 7.00.
Since NH3 is a weak base, it will not be completely consumed in the reaction. However, assuming complete reaction for simplicity, the moles of NH3 needed will be equal to the moles of H3O+ ions. Therefore, you will need 0.225 moles of NH3.
Finally, calculate the volume of 10.0M NH3 solution required to provide 0.225 moles of NH3:
volume (L) = moles ÷ concentration (M)
volume (L) = 0.225 moles ÷ 10.0M
volume (L) = 0.0225 L
Since 1 drop is approximately 0.05 mL, you can calculate the number of drops needed:
drops = volume (L) × (1000 mL / 0.05 mL)
drops = 0.0225 L × (1000 mL / 0.05 mL)
drops = 450 mL
Therefore, you would need approximately 450 drops of 10.0M NH3 solution to raise the pH of the NH4Cl solution to 7.00.