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?
I would use the NH3 since it would be so easy to overshoot with HCl.
How much NH3? Two approaches, both the same but a different look.
NH3 + HOH ==> NH4^+ + OH^-
Then Kb = (NH4^+)(OH^-)/(NH3)
You know Kb, NH4^+, and OH^- so you can calculate molarity of NH3. From there you go to moles NH3 and from there to drops of the 10.0 M solution to add.
The other approach, I think, is more complicated, but you can use the Henderson-Hasselbalch equation to calculate the base needed; the remainder of the problem is the same as the first one.
can you show calculation plz
how many grams does 90 degree to the third power of aluminum weight if it's density is 2.7cm to the third power is equal V=?
To determine which solution to use, we need to consider the chemical reaction that occurs when NH4Cl is dissolved in water. NH4Cl is a salt formed by the combination of ammonium (NH4+) and chloride (Cl-) ions. When NH4Cl is dissolved in water, it dissociates into NH4+ and Cl- ions.
NH4Cl(s) -> NH4+(aq) + Cl-(aq)
Since we want to raise the pH of the solution to 7.00, we need to add a base (or a weakly basic solution) to neutralize the acidic NH4+ ions and raise the pH. In this case, we would use NH3 (ammonia), which is a weak base.
NH3(aq) + H2O(l) -> NH4+(aq) + OH-(aq)
Therefore, we would use the 10.0M NH3 solution.
Now, let's move on to part b of the question to determine how many drops of the 10.0M NH3 solution should be added to reach a pH of 7.00.
To calculate the number of drops needed, we need to determine the volume of the 10.0M NH3 solution required.
First, we need to know the initial concentration of NH4+ ions. Since we have 0.500L of 0.450M NH4Cl solution, we can calculate the moles of NH4+ ions present:
moles of NH4+ ions = volume of solution (L) × concentration of NH4+ ions (M)
moles of NH4+ ions = 0.500L × 0.450M = 0.225 mol
Since the NH4+ ions and NH3 undergo a 1:1 mole ratio reaction, we need to add 0.225 mol of NH3 to reach a pH of 7.00.
Next, we can determine the volume of the 10.0M NH3 solution needed using its concentration:
volume of 10.0M NH3 solution (L) = moles of NH3 / concentration of NH3 (M)
volume of 10.0M NH3 solution = 0.225 mol / 10.0M = 0.0225 L
Now, we know that 1 drop is equivalent to 0.05 mL. To find the number of drops needed, we can convert the volume of the 10.0M NH3 solution from liters to milliliters:
volume of 10.0M NH3 solution in mL = volume of 10.0M NH3 solution (L) × 1000 mL/L
volume of 10.0M NH3 solution in mL = 0.0225 L × 1000 mL/L = 22.5 mL
Finally, we divide the volume in milliliters by the volume per drop to find the number of drops needed:
number of drops = volume of 10.0M NH3 solution in mL / volume per drop
number of drops = 22.5 mL / 0.05 mL/drop = 450 drops
Therefore, you would need 450 drops of the 10.0M NH3 solution to bring the pH of the 0.500L of 0.450M NH4Cl solution to 7.00.