0.194M NaOH and 0.268 M NH4CL Find concentration of OH
I would try this. Set up a RICE chart
R......OH^- + NH4^+ ==> NH3 + H2O\
I...0.194.....0.268......0......0
C..-0.194... -0.194....+0.194.....
E.....0.......0.074.....0.194
This ends up being a buffer of NH4Cl/NH3. All of the NaOH is used.
The Henderson-Hasselbalch equation will solve for pH and convert that to OH^-
pH = pKa + log(NH3(/(NH4^+)
Thanks got it!
To find the concentration of OH- in the solution, we need to understand the concept of acid-base reactions.
NaOH is a strong base, while NH4Cl is a salt that contains the ammonium ion (NH4+) and the chloride ion (Cl-). NH4Cl does not directly contribute to the concentration of OH- in the solution.
However, in solution, NH4Cl will undergo hydrolysis to produce NH4+ and OH- ions. This hydrolysis reaction can be represented as follows:
NH4Cl + H2O -> NH4+ + Cl- + H2O
-> NH4+ + OH- + HCl
From the equation above, we can see that for every one mole of NH4Cl that hydrolyzes, one mole of OH- is produced.
Given the concentrations of NaOH and NH4Cl, let's assume that we have 1 liter of the solution. Therefore, the moles of OH- produced by NH4Cl can be calculated as follows:
Moles of OH- = Moles of NH4Cl
Moles of NH4Cl = concentration of NH4Cl (M) x volume of solution (L)
Moles of NH4Cl = 0.268 M x 1 L = 0.268 moles
Therefore, the concentration of OH- in the solution is 0.268 M.
Note: This calculation assumes that the volume of the solution is 1 liter. If the volume of the solution is different, you can simply multiply the concentration by the volume to obtain the moles of NH4Cl.