a solution containing 100.0 mL of 0.135 M CH3COOH (ka=1.8E-5) is being titrated with 0.54 M NaOH. calculate the ph:
at the equivalence point and 5 mL past the equivalence point.
how much is 5 mL past? i do not know how to do it?
For the equivalence point, I've already posted steps on how to work on it on the other chemistry problem you've posted. So here, I'll just guide you on the "5 mL past equivalence point".
To determine the volume of the point 5 mL past the equivalence point, first, we get the volume of 0.54 M NaOH that was added at equivalence point:
0.135 M * 100 mL = 13.5 mmol CH3COOH
At the balanced reaction, their mole ratio is 1:1, thus,
13.5 mmol CH3COOH * (1 mmol NaOH / 1 mmol CH3COOH) = 13.5 mmol NaOH
M = n/V
V = n/M
V,NaOH = 13.5 mmmol / 0.54 M
V,NaOH = 25 mL
At equivalence point, the total volume is 125 mL.
5 mL past the equivalence point means 125 + 5 = 130 mL. That 5 mL is the 0.54 M NaOH.
At this point, there is excess NaOH (or OH- ions). Getting the moles of OH-,
M = n/V
n = MV
n = 0.54 M * 5 mL
n = 2.7 mmol OH- (excess)
Solve for the concentration:
M = n/V
M = 2.7 mmol / 130 mL
M = 0.020769 M OH-
pOH = -log(OH-)
pOH = -log(0.020769)
pOH = 1.683
pH = 14 - pOH
pH = 12.317
hope this helps~ `u`
I STILL NEED HELP WITH EQUIVALENCE POINT WITH THIS PROBLEM
thanks
Worcester state:
pH- log[H-]<--H20<--> H + 0H-
To calculate the pH at the equivalence point and 5 mL past the equivalence point, we will use the concept of moles of acid and base reacting.
1. Determining the equivalence point:
At the equivalence point, the moles of added NaOH will be equal to the moles of CH3COOH initially present in the solution. This means that the number of moles of CH3COOH will be completely neutralized by the same amount of moles of NaOH.
To find the number of moles of CH3COOH initially present, we can use the formula:
moles = concentration (M) × volume (L)
moles of CH3COOH = 0.135 M × 0.100 L (the given solution volume is 100.0 mL, which is equal to 0.100 L) = 0.0135 moles
Since NaOH is a strong base, it completely dissociates into Na+ and OH- ions. Thus, for every mole of NaOH, we get one mole of OH- ions.
As CH3COOH is a weak acid, it only partially dissociates into CH3COO- and H+ ions. The molar ratio of CH3COOH and H+ ions depends on the acid's dissociation constant (Ka), which is given as 1.8 × 10^-5. To determine the moles of H+ ions, we will use the equation for the acid dissociation:
CH3COOH ⇌ CH3COO- + H+
Using the Henderson-Hasselbalch equation, we can express the concentration of H+ ions at the equivalence point:
[H+] = Ka × (CH3COOH/CH3COO-)
[H+] = 1.8 × 10^-5 × (0.0135 moles / 0.0135 moles) = 1.8 × 10^-5 moles
To find the pH, we can use the equation:
pH = -log[H+]
pH = -log(1.8 × 10^-5) ≈ 4.75
Therefore, at the equivalence point, the pH is approximately 4.75.
2. Determining the pH 5 mL past the equivalence point:
To determine the pH 5 mL past the equivalence point, we need to consider the excess moles of NaOH added.
The 5 mL volume corresponds to an additional 0.005 L. So, the number of moles of NaOH added would be:
moles of NaOH = 0.54 M × 0.005 L = 0.0027 moles
Since NaOH is a strong base that completely dissociates, these moles of NaOH will react with the remaining moles of CH3COOH.
However, the concentration of CH3COOH decreases due to the reaction, resulting in a smaller fraction of dissociation. In this case, we can use the Henderson-Hasselbalch equation again to calculate the pH.
Using the moles of CH3COOH remaining after neutralization:
moles of CH3COOH remaining = 0.0135 moles - 0.0027 moles = 0.0108 moles
[H+] = Ka × (moles of CH3COOH remaining / moles of CH3COO-)
[H+] = 1.8 × 10^-5 × (0.0108 moles / 0.0135 moles) = 1.445 × 10^-5 moles
pH = -log(1.445 × 10^-5) ≈ 4.84
Therefore, 5 mL past the equivalence point, the pH is approximately 4.84.
In summary:
- At the equivalence point, the pH is approximately 4.75.
- 5 mL past the equivalence point, the pH is approximately 4.84.