What is the pH of the solution if 0.015 mol of HCl is added to a buffer containing 0.014 mol of RCOOH and 0.033 mol of RCOONa (source of RCOO-)? Ka = 2.8E-5
I have tried the following:
adding 0.014 + 0.015 = 0.029
ph = pka + log (Base/Acid)
pH = -log(2.8E-5) + log(0.033/0.029)
pH = 4.61
pH = pka + log (Base/Acid)
pH = -log(2.8E-5) + log(0.033/0.014)
pH = 4.93
(i was desparate, i knew that this wasn't the answer)
i have tried to do and ICE diagram, where x ends up = 2.459E-5.
add that to RCOO- to equal 0.03302459, RCOOH = 0.029 (acid)
pH = -log(2.8E-5) + log(0.03302459/0.029)
pH = 4.61
Not sure what to do. Please help
Lynn
ICE is the way to go.
Technically, one should use concentrations but in situations like this the volume (M = mols/L) cancels and we can use mols directly. The equation is and we use mols instead of molarity. Some profs will count off if you don't use M. If your prof does that, then everywhere you plug in mols into the Henderson-Hasselbalch equation, just plug in mol/v, the v cancels, and you go from there.
........RCOO^- + H^+ ==> RCOOH
I......0.033.....0.......0.014
add............0.015........
C....-0.015...-0.015....+0.015...
E.......?........0.......0.015
Now plug the E line into the H-H equation and solve for pH.
To solve this problem, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by pH = pKa + log(Base/Acid), where pKa is the negative logarithm of the acid dissociation constant (Ka), Base is the concentration of the conjugate base, and Acid is the concentration of the acid.
Firstly, let's determine the concentrations of the conjugate base and acid after the addition of HCl. We are given that the initial concentration of RCOOH is 0.014 mol and the initial concentration of RCOONa is 0.033 mol.
After adding 0.015 mol of HCl, the concentration of RCOOH will decrease by 0.015 mol, so it becomes 0.014 mol - 0.015 mol = -0.001 mol. Since the concentration cannot be negative, we can consider it to be zero.
The concentration of RCOONa does not change since it is a salt and does not react with HCl.
Now, we can substitute these values into the Henderson-Hasselbalch equation:
pH = pKa + log(Base/Acid)
Given:
pKa = -log(Ka) = -log(2.8E-5)
Base = RCOONa concentration = 0.033 mol
Acid = RCOOH concentration = 0 mol (considered as 0 due to negative value)
pH = -log(2.8E-5) + log(0.033/0)
pH = -log(2.8E-5)
pH ≈ 4.55
Therefore, the pH of the solution after adding 0.015 mol of HCl is approximately 4.55.