A 50 ml sample of .150M Ch3CooH is titrations with .150 M NaOh solution. Calculate the ph after the following volumes of base have been added: 49 ml, 51ml
CH3COOH + NaOH --> CH3COONa + H2O
millimols CH3COOH = 50 x 0.150 = ?
millimols NaOH = 49 x 0.150 = ?
mmols CH3COOH left = the difference.
M CH3COOH = mmols/total mL = ?
Then convert to pH.
for 49 ml: pH=2.8
for 51 ml: pH=11.17
To calculate the pH after each volume of NaOH solution has been added, we need to consider the reaction that occurs between acetic acid (CH3COOH) and sodium hydroxide (NaOH). The reaction between these two compounds produces sodium acetate (CH3COONa) and water (H2O).
The balanced equation for the reaction is as follows:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, we can see that for every one mole of CH3COOH, one mole of NaOH reacts to produce one mole of CH3COONa and one mole of water.
Now, let's calculate the number of moles of acetic acid present in the 50 ml sample of 0.150 M CH3COOH:
Molarity (M) = moles/volume (L)
0.150 M = moles/0.050 L
moles = 0.150 M x 0.050 L = 0.0075 moles
Since we have excess NaOH solution (0.150 M), we can assume that all the moles of acetic acid will react and be converted into moles of sodium acetate. Therefore, the number of moles of sodium acetate formed will also be 0.0075 moles.
To calculate the pH after adding 49 ml of the NaOH solution:
First, calculate the number of moles of NaOH added:
Molarity (M) = moles/volume (L)
0.150 M = moles/0.049 L
moles = 0.150 M x 0.049 L = 0.00735 moles
Now, let's consider the reaction between CH3COOH and NaOH. Since the stoichiometry between CH3COOH and NaOH is 1:1, adding 0.00735 moles of NaOH will consume 0.00735 moles of CH3COOH.
The remaining moles of CH3COOH can be calculated:
moles of CH3COOH remaining = initial moles of CH3COOH - moles of NaOH added
moles of CH3COOH remaining = 0.0075 moles - 0.00735 moles = 0.00015 moles
To calculate the new concentration of CH3COOH after adding 49 ml of NaOH, we need to consider the new volume of the solution:
New volume = initial volume - volume of NaOH added
New volume = 50 ml - 49 ml = 1 ml = 0.001 L
The new concentration of CH3COOH can be calculated:
New concentration (M) = moles of CH3COOH remaining / new volume (L)
New concentration (M) = 0.00015 moles / 0.001 L = 0.15 M
The pH of the solution can be calculated using the Ka value for acetic acid, which is 1.8 × 10^-5.
pH = -log[H+]
To calculate [H+], we need to recognize that acetic acid is a weak acid, so we can assume it dissociates incompletely. Therefore, we need to calculate the concentration of the dissociated hydrogen ions ([H+]).
[H+] = square root of (Ka x concentration of CH3COOH)
[H+] = square root of (1.8 × 10^-5 x 0.15)
[H+] = square root of (2.7 × 10^-6)
[H+] = 5.2 × 10^-3
Taking the negative logarithm of [H+] yields the pH:
pH = -log(5.2 × 10^-3)
pH = 2.28 (rounded to two decimal places)
To calculate the pH after adding 51 ml of NaOH:
Repeat the calculation by subtracting the moles of NaOH added (calculated as 0.00735 moles) from the initial moles of CH3COOH (0.0075 moles) and calculating the new concentration of CH3COOH based on the new volume of the solution (1 ml):
Moles of CH3COOH remaining = 0.0075 moles - 0.00735 moles = 0.00015 moles
New concentration (M) = 0.00015 moles / 0.001 L = 0.15 M
Calculate [H+] using the new concentration of CH3COOH:
[H+] = square root of (1.8 × 10^-5 x 0.15)
[H+] = square root of (2.7 × 10^-6)
[H+] = 5.2 × 10^-3
Calculate pH:
pH = -log(5.2 × 10^-3)
pH = 2.28 (rounded to two decimal places)
Therefore, the pH after adding 49 ml or 51 ml of NaOH solution is 2.28.