calculate the pH in a solution prepared by dissolving .10 mol of solid NH4Cl in .500 L of .40 M NH3. Assuming no volume change.
This is a buffer problem.
Use the Henderson-Hasselbalch equation.
calculate the pH in a solution prepared by dissolving .10 mol of solid NH4Cl in .500 L of .40 M NH3. Assuming no volume change. Without the use of Henderson-hasselbalch
The HH equation is much easier but here is the other.
(NH3) = 0.40M
(NH4Cl) = (NH4^+) = (Cl^-) = mols/L = 0.1mol/0.500L = 0.2M
...........NH3 + HOH ==> NH4^+ + OH^-
I........0.40..........0.20.......0
C..........-x.............x.......x
E.........0.40-x.......9,.20+x....x
Kb = (NH4^+)(OH^-)/(NH3)
Substitute the E line into Kb above and solve for x = (OH^-). Convert that to pH.
Note: The NH4Cl is a common ion to the NH3 solution (the NH4^+ is the common ion) and the addition of the NH4^+ forces the ionization due to Kb to the left and that limits the basicity. You can assume, to make the solution easier, that 0.200+x = 0.200 and 0.4-x = 0.4.
Well, well, well, pH calculations, huh? Hold on to your funny bone, because I'm about to clown around with some chemistry!
To solve this, we need to follow a little pH dance routine. First, we find the concentration of NH4+ ions created by dissolving the solid NH4Cl. Since 0.10 mol of NH4Cl was dissolved in 0.500 L, we have a concentration of 0.10 mol/0.500 L = 0.20 M for the NH4+ ions.
Now, here comes our lovely assistant, the NH3! With a concentration of 0.40 M, it's ready to tango with the NH4+ ions. NH3 is a basic compound that will accept protons (H+ ions), creating the ammonium ion (NH4+).
The reaction looks like this:
NH3 + H2O ↔ NH4+ + OH-
Based on this reaction, we can deduce that the concentration of OH- ions will be equal to the concentration of NH4+ ions. So, in this case, we have a concentration of 0.20 M for both NH4+ and OH- ions.
Now, let's go backstage and find out how to calculate the pOH. Oh, wait, I forgot to tell you that pOH = -log[OH-]. So, pOH = -log(0.20) = 0.70.
But our true star is the pH, so let's bring it on stage! The pH is determined by the equation pH + pOH = 14. Doing the math, we have pH + 0.70 = 14, which means that pH = 14 - 0.70 = 13.30.
Ta-da! The pH of the solution is approximately 13.30. I hope my chemistry comedy routine brightened up your day!
To calculate the pH of a solution, we need to consider the dissociation of NH4Cl in water and the subsequent reaction with NH3. Here's how you can calculate the pH in this scenario:
Step 1: Write the balanced equation for the dissociation of NH4Cl in water:
NH4Cl(s) ⇌ NH4+(aq) + Cl-(aq)
Step 2: Calculate the concentration of NH4+ ions. Since the solution is prepared by dissolving 0.10 moles of NH4Cl in 0.500 L of water, the concentration of NH4+ ions is:
[(0.10 moles) / (0.500 L)] = 0.20 M
Step 3: Consider the reaction between NH4+ and NH3:
NH4+(aq) + NH3(aq) ⇌ NH4OH(aq)
NH4OH is a weak base that will dissociate to form NH4+ and OH- ions.
Step 4: Calculate the concentration of NH3. The initial concentration of NH3 is 0.40 M, and since it's a weak base, it doesn't dissociate significantly. Therefore, we can assume that the concentration of NH3 remains constant.
Step 5: Calculate the concentration of OH- ions. For every NH4+ ion that reacts with NH3, one NH4OH molecule is formed, which dissociates to form one OH- ion. Hence, the concentration of OH- ions is equal to the concentration of NH4+ ions, which is 0.20 M.
Step 6: Calculate the pOH. The pOH is directly related to the concentration of OH- ions by the equation:
pOH = -log[OH-]
pOH = -log(0.20) ≈ 0.70
Step 7: Calculate the pH. The pH is related to pOH by the equation:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 0.70 ≈ 13.30
Therefore, the pH of the solution prepared by dissolving 0.10 mol of NH4Cl in 0.500 L of 0.40 M NH3 would be approximately 13.30.