The pH of a solution prepared by mixing 55.0 mL of 0.183 M KOH and 10.0 mL of 0.145 M HC2H3O2 is ________.
the answer is 13.122 however when i try to plug it into the Henderson Hasselbalch equation (pH= -log(1.8E-5)+log( 0.010065/0.00145)
i am not getting 13.122
the limiting reagent i think it is HC2H3O2
55.0 mL (1L/1000mL)(0.183mol/1L) = 0.010065 mol KOH
10.0 mL (1L/1000mL)(0.145mol/1L) = 0.00145 mol HC2H3O2
I do not know what I am doing wrong to get a pH of 5.58 instead of 13.122
can anyone tell me what I am doing wrong? I appreciate any help.
Thank You!
Yes, you are assuming you have buffered solution but you don't. You have an excess of KOH, which you note when you say acetic acid is the limiting regent. So (OH^-) = mols KOH/L solution. Solve for pOH and convert to pH. 13.12
To find the pH of the solution, you need to use the Henderson-Hasselbalch equation. However, there seems to be a mistake in the equation you provided.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, the acid is HC2H3O2 (acetic acid) and the conjugate base is C2H3O2- (acetate ion). The Ka value for acetic acid is 1.8 x 10^-5.
To use the Henderson-Hasselbalch equation, you need to evaluate the concentrations of the acid and the conjugate base.
First, let's calculate the concentration of acetic acid (HA):
10.0 mL (1L/1000mL)(0.145 mol/L) = 0.00145 mol
Next, let's calculate the concentration of acetate ion (A-):
55.0 mL (1L/1000mL)(0.183 mol/L) = 0.010065 mol
Now, let's plug these values into the Henderson-Hasselbalch equation:
pH = -log(1.8 x 10^-5) + log(0.010065/0.00145)
Using a calculator, we can solve this equation:
pH ≈ 4.74
Therefore, the pH of the solution is approximately 4.74, not 13.122.
It appears that there was an error in calculating the pH using the Henderson-Hasselbalch equation. Double-check your calculations and ensure that you are using the correct values for the concentrations and the pKa value.
To find the pH of a solution prepared by mixing two solutions, you need to consider the reaction that occurs between the two solutes. In this case, KOH (a strong base) reacts with HC2H3O2 (a weak acid) to form water and a salt, which is KC2H3O2.
The balanced equation for this reaction is:
HC2H3O2 + KOH -> KC2H3O2 + H2O
Since KOH is a strong base, it completely dissociates in water. HC2H3O2, on the other hand, is a weak acid, so it does not fully dissociate. You can use the Henderson-Hasselbalch equation to calculate the pH of the resulting solution:
pH = pKa + log ([A-]/[HA])
In this equation, pKa is the negative logarithm of the acid dissociation constant of the weak acid (HC2H3O2), [A-] is the concentration of the conjugate base (C2H3O2-), and [HA] is the concentration of the weak acid (HC2H3O2) remaining in the solution.
First, you need to calculate the concentrations of [A-] and [HA].
For [A-]:
Calculate moles of C2H3O2-:
0.00145 mol HC2H3O2 x (1 mol C2H3O2-/1 mol HC2H3O2) = 0.00145 mol C2H3O2-
Calculate concentration of C2H3O2-:
0.00145 mol C2H3O2- / (55 mL + 10 mL) = 0.00145 mol/65 mL = 0.0223 M
For [HA]:
Calculate moles of HC2H3O2 that did not react:
0.145 mol HC2H3O2 - 0.00145 mol HC2H3O2 = 0.14355 mol HC2H3O2
Calculate concentration of HC2H3O2:
0.14355 mol HC2H3O2 / (55 mL + 10 mL) = 0.14355 mol/65 mL = 0.0221 M
Next, you need to determine the pKa value for HC2H3O2. The pKa of acetic acid (HC2H3O2) is around 4.76.
Now you can substitute the values into the Henderson-Hasselbalch equation:
pH = 4.76 + log (0.0223/0.0221)
= 4.76 + log (1.009)
= 4.76 + 0.004
= 4.764
Therefore, the pH of the solution is 4.764, which is different from the given answer of 13.122. It appears that you have made an error in the calculation or data input somewhere. Double-check your calculations to ensure accuracy.