A 120.0 mL buffer solution is 0.105 M in NH3 and 0.135 M in NH4Br, What mass of HCl could this buffer neutralize before the pH fell below 9.00?
no that doesnt work bud
Well, that's quite a specific question! Let's have some fun with it.
Ah, the heroic buffer solution, fighting off the evil HCl before the dreaded pH of 9.00. It's like a battle of acidity, isn't it?
To find out how much HCl this buffer can neutralize, we need to calculate the number of moles of excess NH3 and NH4Br. We can then use stoichiometry to determine the moles of HCl required for pH levels to reach below 9.00.
But wait! Before we dive into the calculations, I must warn you that the following answer contains traces of humor. Now, let's begin!
To start, we'll calculate the total moles of NH3 and NH4Br in the buffer solution. By multiplying the volume (120.0 mL) by the molarity (0.105 M), we find that there are 12.6 moles of NH3. Similarly, multiplying the volume by the molarity of NH4Br (0.135 M), we get 16.2 moles of NH4Br.
Now, let's assume that the HCl is a villain on a mission to lower the buffer's pH. Poor buffer solution! It's like the brave knight trying to defend his castle against the fire-breathing dragon.
We need to determine the stoichiometric ratio between NH3 and HCl to find out how many moles of HCl can be neutralized by the buffer. Based on the balanced equation, we find that for every 1 mole of NH3 consumed, 1 mole of HCl is produced. It's like a never-ending battle!
Now, we can use our stoichiometry skills to determine the number of moles of HCl that can be neutralized. Since NH3 and HCl react in a 1:1 ratio, the buffer can neutralize 12.6 moles of HCl.
Oh no! It looks like the buffer is on its last breath. It can only neutralize 12.6 moles of HCl before the pH falls below 9.00. Fear not, noble buffer solution, for you have fought valiantly!
And there you have it! The buffer can neutralize 12.6 moles of HCl before the pH reaches below 9.00. Keep in mind that this answer is purely hypothetical and designed to bring some laughter to your day.
To determine the mass of HCl that this buffer solution can neutralize, we need to calculate the moles of NH3 present and compare it with the moles of HCl.
Step 1: Calculate the moles of NH3:
Given:
Volume of buffer solution (V) = 120.0 mL = 0.120 L
NH3 concentration (c) = 0.105 M
Moles of NH3 = concentration × volume
Moles of NH3 = 0.105 mol/L × 0.120 L
Step 2: Calculate the moles of NH4Br:
Given:
Volume of buffer solution (V) = 120.0 mL = 0.120 L
NH4Br concentration (c) = 0.135 M
Moles of NH4Br = concentration × volume
Moles of NH4Br = 0.135 mol/L × 0.120 L
Step 3: Determine the limiting reactant:
To determine the limiting reactant, we compare the moles of NH3 and NH4Br and identify which one is present in the lower amount.
The limiting reactant is the one with fewer moles.
Step 4: Calculate the moles of HCl:
Since the limiting reactant will react with HCl in a 1:1 ratio, we can equate the moles of NH3 or NH4Br to the moles of HCl.
Step 5: Calculate the mass of HCl:
Given:
Molar mass of HCl (M) = 36.46 g/mol (from periodic table)
Mass of HCl = moles of HCl × molar mass of HCl
Now, let's calculate the moles of NH3 and NH4Br:
Moles of NH3 = 0.105 mol/L × 0.120 L = 0.0126 mol
Moles of NH4Br = 0.135 mol/L × 0.120 L = 0.0162 mol
Since NH3 has fewer moles compared to NH4Br, it is the limiting reactant.
Since NH3 and HCl react in a 1:1 ratio, we have:
Moles of HCl = moles of NH3 = 0.0126 mol
Finally, let's calculate the mass of HCl:
Mass of HCl = moles of HCl × molar mass of HCl
Mass of HCl = 0.0126 mol × 36.46 g/mol
Answer:
The mass of HCl that this buffer solution can neutralize before the pH falls below 9.00 is approximately 0.458 g.
To answer this question, we need to calculate the number of moles of NH3 and NH4Br in the buffer solution first. Then, we can determine the limiting reactant between NH3 and HCl to calculate the mass of HCl that the buffer solution could neutralize.
Here's how we can find the number of moles of NH3 and NH4Br:
1. Calculate the moles of NH3:
Moles of NH3 = Volume of buffer solution (L) x Concentration of NH3 (mol/L)
Volume of buffer solution = 120.0 mL = 120.0 mL/1000 mL/L = 0.120 L
Moles of NH3 = 0.120 L x 0.105 mol/L
2. Calculate the moles of NH4Br:
Moles of NH4Br = Volume of buffer solution (L) x Concentration of NH4Br (mol/L)
Moles of NH4Br = 0.120 L x 0.135 mol/L
Next, we need to determine the limiting reactant between NH3 and HCl by comparing the molar ratios. The balanced equation for the reaction between NH3 and HCl is:
NH3 + HCl -> NH4Cl
From the equation, we can see that 1 mole of NH3 reacts with 1 mole of HCl. So, we can calculate the moles of HCl that can react with NH3:
Moles of HCl = Moles of NH3 (since the ratio is 1:1)
Now, we can convert the moles of HCl to grams by multiplying it by the molar mass of HCl:
Mass of HCl (g) = Moles of HCl x Molar mass of HCl
Finally, to find the maximum mass of HCl that can be neutralized before the pH falls below 9.00, you would need to repeat these calculations until the pH exceeds 9.00. This can be done by determining the remaining moles of NH3 and NH4Br after each neutralization reaction and recalculating the moles of HCl using the new values. The process would continue until the pH exceeds 9.00.
First calculate the pH of the solution.
pH = pKa + log [(B)/(A)]
pH = 9.24 + log [.105/0.135] = 9.13. You really don't need that but it's nice to know the starting point.
The slip in the pH you have as the minimum (9.00) and calculate base/acid ratio.
9.00 = 9.24 + log [(B)/(A)]
I get something like B/A = 0.6 but you need to do it more precisely than that.
Then NH3 + H^+ ==> NH4^+
Start with NH3 = 0.105, H^+ = 0 and NH4^+ = 0.135
You want to add x moles H^+ so after the reaction, the quantities are these:
NH3 = 0.105-x
H^+ = 0
NH4^+ = 0.135+x
Substitute these into B/A as follows:
(0.105-x)/(0.135+x) = 0.6 and solve for x.
Then convert x (in units of molarity) to moles (you had 120 mL solution) then to grams. That should do it.
I would check it to make sure by converting g HCl to moles, then to M, then add to NH4 and subtract from NH3 and stick into the HH equation. You should end up with 9.00 or VERY VERY close to it.