Calculate the pH of a solution obtained by mixing 30 mL of 0.75 M CH3COOH with 15 mL of 1.5 M NaOH.
millimoles NaOH = mL x M = 15 x 1.5 = 22.5
mmoles CH3COOH = 30 x 0.75 = 22.5 where CH3COOH = HAc.
.........HAc + NaOH ==> NaAc + H2O
initial..22.5...22.5.....0......0
change..-22.5..-22.5....+22.5..22.5
equil.....0......0.......22.5....
So you can see that the CH3COOH is exactly neutralized by the NaOH and neither free CH3COOH nor free NaOH come through the reaction. We have just a solution of CH3COONa (NaAc) and that is hydrolyzed. The pH is determined by that. (Ac^-) = mmoles/mL = 22.5/45 = 0.5M
..........Ac^- + HOH ==> HAc + OH^-
initial..0.50....0.......0.....0
change.....-x..............x......x
equil......0.50-x..........x.......x
Kb for Ac^- = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)
Kw = 1E-14 for water.
Ka = 1.8E-5 for acetic acid
x = (HAc) = (OH^-)
(Ac^-) = 0.50-x
Substitute, solve for x (which = OH^-), convert to pOH, then to pH.
gibbe that answer b0ss
To calculate the pH of the solution obtained by mixing CH3COOH and NaOH, we need to determine the concentration of the resulting solution and then calculate the pH.
Step 1: Determine the limiting reactant
To determine the limiting reactant, we need to calculate how many moles of each reactant are present.
For CH3COOH:
Volume (V) = 30 mL = 0.03 L
Molarity (M) = 0.75 M
Number of moles (n) = M x V = 0.75 M x 0.03 L = 0.0225 moles
For NaOH:
Volume (V) = 15 mL = 0.015 L
Molarity (M) = 1.5 M
Number of moles (n) = M x V = 1.5 M x 0.015 L = 0.0225 moles
Since the number of moles of CH3COOH and NaOH are equal, there is no limiting reactant in this case.
Step 2: Calculate the concentration of the resulting solution
When CH3COOH and NaOH react, they form water (H2O) and sodium acetate (CH3COONa).
The balanced chemical equation for the reaction is:
CH3COOH + NaOH -> CH3COONa + H2O
Since the number of moles of CH3COOH and NaOH is equal, they will react in a 1:1 mole ratio. This means that all of the CH3COOH will react with NaOH, resulting in the formation of sodium acetate and water.
The volume of the resulting solution is the sum of the volumes of CH3COOH and NaOH:
Vresulting = VCH3COOH + VNaOH = 0.03 L + 0.015 L = 0.045 L
To find the new concentration of the resulting solution, divide the total number of moles by the new volume:
nresulting = nCH3COOH + nNaOH = 0.0225 moles + 0.0225 moles = 0.045 moles
Cresulting = nresulting / Vresulting = 0.045 moles / 0.045 L = 1 M (because the volume doubled)
Step 3: Calculate the pH of the resulting solution
To calculate the pH, we need the concentration of H+ ions, which is determined by the concentration of CH3COOH and the dissociation constant (Ka) of CH3COOH.
The dissociation constant (Ka) of CH3COOH is 1.8 x 10^-5.
Using the equation for the dissociation of CH3COOH:
CH3COOH -> CH3COO- + H+
The initial concentration of CH3COOH is 0.75 M, and assuming complete dissociation, the concentration of H+ ions is also 0.75 M.
The pH can be calculated using the equation:
pH = -log[H+]
So, for the resulting solution, with a concentration of H+ ions of 1 M:
pH = -log(1) = 0
Therefore, the pH of the solution obtained by mixing 30 mL of 0.75 M CH3COOH with 15 mL of 1.5 M NaOH is 0.
To calculate the pH of the solution obtained by mixing CH3COOH (acetic acid) with NaOH (sodium hydroxide), you need to determine the concentration of the resulting species, particularly H+ or OH- ions.
The balanced chemical equation for the reaction between CH3COOH and NaOH is:
CH3COOH + NaOH -> CH3COONa + H2O
Since acetic acid is a weak acid, it does not completely dissociate in water, but partially ionizes. However, sodium hydroxide is a strong base and fully dissociates into sodium (Na+) and hydroxide (OH-) ions when dissolved in water.
Step 1: Calculate the number of moles of CH3COOH and NaOH.
Moles of CH3COOH = volume (in liters) x concentration (in moles per liter)
= 0.03 L x 0.75 mol/L
= 0.0225 mol
Moles of NaOH = volume (in liters) x concentration (in moles per liter)
= 0.015 L x 1.5 mol/L
= 0.0225 mol
Step 2: Determine the limiting reactant.
In this case, both reactants have the same number of moles, which means they are present in stoichiometrically equivalent amounts. Therefore, the limiting reactant is either CH3COOH or NaOH as they will react fully, leaving no excess of either.
Step 3: Determine the number of moles of H+ and OH- ions formed.
Since CH3COOH is a weak acid, it partially dissociates:
CH3COOH -> CH3COO- + H+
From the balanced equation, we can see that the stoichiometric relationship between CH3COOH and H+ ions is 1:1. Therefore, the number of moles of H+ ions formed is also 0.0225 mol.
In the reaction between NaOH and H+, each molecule of NaOH releases one hydroxide (OH-) ion. Hence, the number of moles of OH- ions formed is 0.0225 mol.
Step 4: Calculate the total volume of the solution.
The total volume of the solution is obtained by adding the volumes of CH3COOH and NaOH:
Total volume = 30 mL + 15 mL = 45 mL = 0.045 L
Step 5: Calculate the concentration of H+ and OH- ions in the final solution.
Concentration (in moles per liter) = number of moles / total volume (in liters)
Concentration of H+ ions = 0.0225 mol / 0.045 L = 0.5 M
Concentration of OH- ions = 0.0225 mol / 0.045 L = 0.5 M
Step 6: Calculate the pH of the solution.
The pH is a measure of the concentration of H+ ions in a solution. It is calculated using the expression:
pH = -log10 [H+]
Taking the negative logarithm (base 10) of the H+ concentration:
pH = -log10 (0.5)
≈ -(-0.301)
≈ 0.301
Therefore, the pH of the solution obtained by mixing 30 mL of 0.75 M CH3COOH with 15 mL of 1.5 M NaOH is approximately 0.301.