50.0 mL of 0.18 M NH3 with 5.0 mL of 0.36 M HBr.
Kb of NH3 is 1.77 x 10-5
Calculate pH
I don't know where to begin
millimols NH3 initially = mL x M = 50 x 0.18 = 9
millimola HBr initially = 5 x 0.36 = 1.8
..............NH3 + HBr ==> NH4Br
I.............9..........1.8...........0
C.........-1.8......-1.8.........+1.8
E............7.2.........0...........1.8
So your solution has 7.2 mmols NH3 and 1.8 mmols of its salt of NH4B4. That is a buffered solution so apply the Henderson-Hasselbalch equation and calculate pH.
pH = pKa + log (base)/(acid)
Convert Kb for NH3 to Ka then to pKa.
M base = M NH3 = mmols/total mL
M acide = M NH4^+ = mmols/totalmols.
Post your work if you get stuck.
E........
7.2mmols/55ml = .1309 NH3
1.8 mmols/ 55ml = .0327 NH4
-log(1.77 x 10-5) = 4.752
14-4.752 = pKa 9.24798
9.24798 + log (.1309/.0327) = 9.8503
is that right^
That looks OK to me but you aren't allowed that many places in your answer. I would round it to 9,85.
To calculate the pH in this scenario, you need to consider the reaction between NH3 (ammonia) and HBr (hydrobromic acid). The reaction between an acid and a base can lead to the formation of a salt and water.
Here's how you can approach this problem:
Step 1: Find the limiting reactant
To determine the limiting reactant, you need to compare the number of moles of NH3 and HBr. Start by calculating the moles of NH3 and HBr using the concentration and volume given in the problem:
Moles of NH3 = concentration of NH3 (0.18 M) * volume of NH3 (50.0 mL)
Moles of HBr = concentration of HBr (0.36 M) * volume of HBr (5.0 mL)
Step 2: Convert the moles to the same unit for comparison
To accurately compare the moles of NH3 and HBr, convert the volume of HBr from milliliters (mL) to liters (L):
Volume of HBr = 5.0 mL = 0.005 L
Step 3: Calculate the mole ratio
Now that you have the moles of NH3 and HBr, determine the mole ratio between NH3 and HBr based on their balanced chemical equation. The balanced equation is:
NH3 + HBr -> NH4Br
The mole ratio between NH3 and HBr is 1:1, meaning one mole of NH3 reacts with one mole of HBr.
Step 4: Determine the limiting reactant
Compare the moles of NH3 and HBr to determine which one is the limiting reactant. The reactant that produces fewer moles of the product (NH4Br) is considered the limiting reactant. If moles of NH3 are less than moles of HBr, NH3 is the limiting reactant. Otherwise, HBr is the limiting reactant.
Step 5: Calculate the moles of the excess reactant
Subtract the moles of the limiting reactant from the corresponding reactant to find out the moles of the excess reactant.
Step 6: Calculate the concentration of NH4+ (ammonium ion)
Since NH3 is a weak base, it reacts with HBr to form NH4+ (ammonium ion). The concentration of NH4+ can be calculated by determining the moles of NH4+ formed and dividing it by the total volume of the solution (sum of the volumes of NH3 and HBr).
Step 7: Calculate the pOH
Use the concentration of NH4+ to find the pOH. The pOH is calculated using the pOH equation:
pOH = -log10 [OH-]
Since NH4+ is the conjugate acid of NH3, the concentration of OH- can be calculated indirectly. The Kb of NH3 can be utilized to find the concentration of OH- by calculating the concentration of NH3 that reacts with water to produce OH-.