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?
I've posted an answer on your previous post on this similar question. Anyway, I'll just repost it here:
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`
thanks
To calculate the pH at the equivalence point and 5 mL past the equivalence point, we first need to determine the volume of NaOH required to reach the equivalence point.
The balanced chemical equation for the reaction between CH3COOH and NaOH is:
CH3COOH + NaOH -> CH3COONa + H2O
From the stoichiometry of the reaction, we can see that 1 mole of NaOH reacts with 1 mole of CH3COOH.
Given that the concentration of CH3COOH is 0.135 M and the volume is 100.0 mL (0.100 L), we can calculate the number of moles of CH3COOH present using the formula:
moles of CH3COOH = concentration of CH3COOH × volume of CH3COOH
= 0.135 M × 0.100 L
= 0.0135 moles of CH3COOH
Since the stoichiometry of the reaction is 1:1, it means that 0.0135 moles of NaOH will be required to reach the equivalence point.
Next, to determine the volume of NaOH required to reach the equivalence point, we can use the formula:
volume of NaOH = moles of NaOH / concentration of NaOH
= 0.0135 moles / 0.54 M
≈ 0.025 L (or 25 mL)
Therefore, the volume of NaOH required to reach the equivalence point is approximately 25 mL.
To calculate the pH at the equivalence point, we need to determine the concentration of the resulting solution after the reaction at the equivalence point. Since the stoichiometry of the reaction is 1:1, the number of moles of CH3COOH will be equal to the number of moles of CH3COONa (the salt formed) at the equivalence point.
The total volume of the solution at the equivalence point is given by the sum of the initial volume of CH3COOH (100.0 mL) and the volume of NaOH added (25 mL), which equals 125 mL or 0.125 L.
The concentration of CH3COOH at the equivalence point can be calculated using the formula:
concentration of CH3COOH = moles of CH3COOH / volume of solution
= 0.0135 moles / 0.125 L
≈ 0.108 M
Now, let's calculate the pH at the equivalence point using the dissociation of CH3COOH:
CH3COOH + H2O -> H3O+ + CH3COO-
The dissociation constant (Ka) for CH3COOH is given as 1.8 x 10^-5.
To calculate the pH, we need to determine the concentration of H3O+ ions (which is equal to the concentration of CH3COOH at the equivalence point) and then take the negative logarithm to find the pH:
[H3O+] = √(Ka × concentration of CH3COOH)
= √(1.8 x 10^-5 × 0.108 M)
≈ 8.916 x 10^-3 M
pH = -log[H3O+]
= -log(8.916 x 10^-3)
≈ 2.05
Therefore, the pH at the equivalence point is approximately 2.05.
To calculate the pH 5 mL past the equivalence point, we need to consider the excess NaOH that has been added. Since NaOH is a strong base and fully dissociates in water, we can determine the concentration of OH- ions in the solution.
Given that 5 mL of 0.54 M NaOH is added, we can calculate the number of moles of OH- ions added using the formula:
moles of OH- = concentration of NaOH × volume of NaOH
= 0.54 M × 0.005 L
= 0.0027 moles of OH-
Since the stoichiometry of the reaction between NaOH and CH3COOH is 1:1, this means that 0.0027 moles of OH- will react with 0.0027 moles of CH3COOH.
To calculate the concentration of CH3COOH remaining, we subtract the moles of CH3COOH that reacted from the initial moles of CH3COOH:
moles of CH3COOH remaining = initial moles of CH3COOH - moles of OH-
= 0.0135 moles - 0.0027 moles
= 0.0108 moles
The remaining volume of the solution is the initial volume (100.0 mL) plus the volume of NaOH added (5 mL), which equals 105 mL or 0.105 L.
The concentration of CH3COOH remaining can be calculated using the formula:
concentration of CH3COOH = moles of CH3COOH remaining / volume of solution
= 0.0108 moles / 0.105 L
≈ 0.103 M
Again, we can calculate the concentration of H3O+ ions using the dissociation of CH3COOH and find the pH using the same method as before:
[H3O+] = √(Ka × concentration of CH3COOH)
= √(1.8 x 10^-5 × 0.103 M)
≈ 6.337 x 10^-3 M
pH = -log[H3O+]
= -log(6.337 x 10^-3)
≈ 2.198
Therefore, the pH 5 mL past the equivalence point is approximately 2.198.
To calculate the pH at the equivalence point and 5 mL past the equivalence point, we need to consider the stoichiometry of the acid-base reaction between CH3COOH (acetic acid) and NaOH.
First, let's calculate the volume of NaOH required to reach the equivalence point. The equation for the reaction between CH3COOH and NaOH is as follows:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that the stoichiometry is 1:1 between CH3COOH and NaOH. This means that the volume of NaOH needed to reach the equivalence point is the same as the volume of CH3COOH.
In this case, the initial volume of CH3COOH solution is 100.0 mL. Therefore, the volume of NaOH required to reach the equivalence point is also 100.0 mL.
Now, let's calculate the pH at the equivalence point:
At the equivalence point, CH3COOH will react completely with NaOH, forming CH3COONa (sodium acetate) and water. The resulting solution will be a sodium acetate solution.
Since sodium acetate is the salt of a weak acid (CH3COOH), it will undergo hydrolysis. The hydrolysis of sodium acetate can be represented by the equation:
CH3COONa + H2O → CH3COOH + NaOH
The hydrolysis of sodium acetate produces acetic acid and hydroxide ions (OH-). The acetic acid acts as a weak acid and donates H+ ions to the solution, while the OH- ions raise the pH.
To calculate the pH at the equivalence point, we need to determine the concentration of the resulting acetic acid after the complete reaction between CH3COOH and NaOH.
Since the volume of CH3COOH solution is 100.0 mL and the molarity of CH3COOH is 0.135 M, the number of moles of CH3COOH in the solution can be calculated using the formula:
moles = volume (L) x molarity
moles = (100.0 mL / 1000 mL/L) x 0.135 M
moles = 0.0135 mol
Since the stoichiometry of the reaction is 1:1 between CH3COOH and CH3COONa, the concentration of CH3COOH in the resulting solution will also be 0.0135 M.
Next, we need to calculate the concentration of OH- ions in the hydrolyzed solution. Since NaOH is a strong base, it will dissociate completely in water, providing a concentration of 0.54 M OH- ions.
Now, we can use the equilibrium expression for the ionization of acetic acid (CH3COOH), which is given by:
Ka = [CH3COO-][H+]/[CH3COOH]
Since the concentration of CH3COOH is the same as the concentration of CH3COOH resulting from the hydrolysis, we can simplify the equation to:
Ka = [CH3COO-][H+]/0.0135
The value of Ka is given as 1.8 x 10^-5.
Since the CH3COO- and H+ ions are produced in a 1:1 ratio during the hydrolysis of sodium acetate, we can assume that:
[CH3COO-] = [H+]
Applying this to the equilibrium expression, we have:
Ka = [H+]^2/0.0135
Simplifying the equation further, we can rearrange to solve for [H+]:
[H+]^2 = Ka * 0.0135
[H+]^2 = (1.8 x 10^-5) * (0.0135)
[H+]^2 = 2.43 x 10^-7
Taking the square root of both sides, we find:
[H+] = 4.93 x 10^-4 M
Now, we can use the definition of pH to calculate the pH at the equivalence point:
pH = -log[H+]
pH = -log(4.93 x 10^-4)
pH = -log(4.93) + log(10^-4)
pH = -0.69 - (-4)
pH = 3.31
Therefore, at the equivalence point, the pH is approximately 3.31.
To calculate the pH 5 mL past the equivalence point, we need to consider the excess NaOH that has been added. Since the initial volume of NaOH required to reach the equivalence point is 100.0 mL, the total volume of NaOH added once we are 5 mL past the equivalence point is 105 mL (100 mL + 5 mL).
At this point, all the acetic acid (CH3COOH) has been neutralized, and the resulting solution contains only the hydrolyzed sodium acetate.
To calculate the pH, we can use the same principles as before. The concentration of hydroxide ions (OH-) in the solution is still 0.54 M, and the concentration of acetic acid (CH3COOH) resulting from the hydrolysis is still 0.0135 M.
Using the same equilibrium expression and calculation methods, we can find the concentration of H+ ions and then calculate the pH.
The resulting pH 5 mL past the equivalence point will be slightly basic due to the excess concentration of OH- ions.
I hope this explanation helps you understand how to calculate the pH at the equivalence point and 5 mL past the equivalence point in an acid-base titration.