What is the ph when 100 mL of 0.1 M NaOH is added to 150 mL of 0.2 M Hac if pka for acetic acid =4.76?
100 mL x 0.1M NaOH = 10 mmoles.
150 mL x 0.2M HAc = 30 mmoles.
............NaOH + HAc ==> NaAc + H2O
initial.....10......30......0......0
change.....-10.....-10.....10.....10
equil........0......20......10....10
Now substitute into the Henderson-Hasselbalch equation and solve for pH.
4.23
To determine the pH when 100 mL of 0.1 M NaOH is added to 150 mL of 0.2 M Hac (acetic acid), we need to consider the reaction between NaOH and Hac.
NaOH is a strong base, and Hac is a weak acid. When NaOH reacts with Hac, it forms water (H2O) and sodium acetate (NaC2H3O2).
The balanced chemical equation for the reaction is:
NaOH + Hac → H2O + NaC2H3O2
First, let's calculate the moles of NaOH and Hac:
Moles of NaOH = Volume (L) × Concentration (M)
= 0.1 L × 0.1 M
= 0.01 moles
Moles of Hac = Volume (L) × Concentration (M)
= 0.15 L × 0.2 M
= 0.03 moles
Next, we need to find the limiting reagent. The Hac will be completely consumed in the reaction because the moles of Hac (0.03 moles) is greater than the moles of NaOH (0.01 moles).
Now, we need to calculate the moles of the remaining Hac after the reaction:
Moles of Hac remaining = Moles of Hac initially - Moles of NaOH reacted
= 0.03 moles - 0.01 moles
= 0.02 moles
Next, we need to calculate the concentration of the remaining Hac:
Concentration of Hac = Moles of Hac remaining / Volume of solution (L)
= 0.02 moles / (0.15 L + 0.1 L)
= 0.02 moles / 0.25 L
= 0.08 M
Now, let's calculate the pOH of the solution using the concentration of the remaining Hac:
pOH = -log10(Concentration of Hac)
= -log10(0.08)
≈ 1.10
Finally, we can calculate the pH using the pOH and the pKa of acetic acid:
pH = 14 - pOH - pKa
= 14 - 1.10 - 4.76
≈ 8.14
Therefore, the pH when 100 mL of 0.1 M NaOH is added to 150 mL of 0.2 M Hac is approximately 8.14.
To determine the pH when 100 mL of 0.1 M NaOH is added to 150 mL of 0.2 M Hac (acetic acid), we need to consider the reaction that takes place between NaOH and Hac.
First, let's calculate the number of moles of NaOH and Hac present in the solutions:
Number of moles of NaOH = volume (in L) × molarity = 0.1 L × 0.1 mol/L = 0.01 mol
Number of moles of Hac = volume (in L) × molarity = 0.15 L × 0.2 mol/L = 0.03 mol
The stoichiometry of the reaction between NaOH and Hac is 1:1, meaning that 1 mole of NaOH reacts with 1 mole of Hac. In this case, since the number of moles of NaOH is less than the number of moles of Hac, NaOH will be completely consumed and Hac will be in excess.
The reaction between NaOH and Hac produces water and sodium acetate (NaAc):
NaOH + Hac → HOH + NaAc
Since Hac is a weak acid and NaOH is a strong base, the reaction will shift predominantly towards the products. Therefore, sodium acetate (NaAc) will be formed, which is a salt of a weak acid and a strong base. In an aqueous solution, NaAc will undergo hydrolysis, resulting in the formation of hydroxide ions (OH-) and acetic acid (Hac):
NaAc + H2O ⇌ Hac + OH-
To determine the pH of this solution, we need to consider the dissociation of acetic acid (Hac). Acetic acid is a weak acid, so it undergoes partial dissociation in water:
Hac ⇌ H+ + Ac-
The equilibrium constant for this dissociation is represented as Ka, and the pKa is the negative logarithm of Ka. In this case, pKa = 4.76, which means that the Ka can be calculated as 10^(-pKa) = 10^(-4.76).
Now, we can set up an ICE table (Initial, Change, Equilibrium) to determine the concentrations of the ions in solution after the hydrolysis of NaAc:
NaAc + H2O ⇌ Hac + OH-
Initial concentrations: 0.03 mol | 0 mol | 0 mol | 0 mol
Change in concentrations: -x mol | +x mol | +x mol | +x mol
Equilibrium concentrations: (0.03 - x) mol | x mol | x mol | x mol
Since NaAc is the salt of a strong base and a weak acid, the hydrolysis reaction will produce OH- ions, resulting in an increase in the concentration of OH-. The concentration of Hac and H+ will be equal to the concentration of OH-.
The concentration of H+ (and OH-) can be expressed using the equation for the dissociation of acetic acid:
Ka = [H+][Ac-] / [Hac]
Substituting the equilibrium concentrations from the ICE table, we have:
10^(-4.76) = x * x / (0.03 - x)
Simplifying the equation, we get:
10^(-4.76) = x^2 / (0.03 - x)
Solving this equation will give us the value of x, which represents the concentration of H+ (and also OH- and Ac-) in mol/L. Once we have this concentration, we can calculate the pH using the equation pH = -log[H+].
Note: The calculation to solve for x involves solving a quadratic equation. The exact solution involves using the quadratic formula or numerical methods.