1)What is the pH of a 0.25 M solution of acetate (CH3COO-) with a pKa of 4.76?
2)If the [OH] of a solution is 2.95 x 10-5M, what is the pH of the solution?
3)Determine the pH at the equivalence point when 25.00 mL of 0.1056 M CH3COOH is titrated with 0.1056 M NaOH.
4)Determine the pH of a solution after 20.00 mL of 0.4963 M HI has been titrated with 12.64 mL of 0.5174 M NaOH.
Btw for number 4 i got ph of 1, not sure if its right..
In a hurried calculation I got 0.9841 for pH and I carried it out to 4 significant figures and used all of the digits in the numbers themselves.
1. acetate = Ac
......Ac + HOH ==> HAc + OH^-
I....0.25..........0......0
C.....-x...........x......x
E...0.25-x.........x......x
Kb for Ac = (Kw/Ka for HAc) = (x)(x)/(0.25-x)
Solve for x = (OH^-) and covert to pH. The answer is around 8+.
1) To determine the pH of a solution of acetate, we can use the Henderson-Hasselbalch equation. The equation is given as:
pH = pKa + log([A-]/[HA])
Here, pKa is the negative logarithm of the acid dissociation constant (Ka) of acetate, [A-] is the concentration of the acetate ion, and [HA] is the concentration of the undissociated acetic acid.
Given that the pKa of acetate is 4.76 and the concentration of the solution is 0.25 M, we need to find the concentrations of [A-] and [HA]. Since acetate is the conjugate base of acetic acid, we assume that it is fully dissociated and [A-] is equal to 0.25 M. Therefore, the concentration of [HA] can be calculated as 0 M (because it is fully dissociated).
Plugging these values into the Henderson-Hasselbalch equation:
pH = 4.76 + log(0.25/0) = 4.76 + log(∞) = 4.76 + ∞ = ∞
Therefore, the pH of the 0.25 M solution of acetate is approximately infinity (∞).
2) To determine the pH of a solution based on the hydroxide ion concentration ([OH]), we can use the following equation:
pH = 14 - log([OH])
Given that [OH] in the solution is 2.95 x 10^(-5) M, we can plug this value into the equation:
pH = 14 - log(2.95 x 10^(-5)) = 14 + (-log(2.95 x 10^(-5)))
Using a calculator, we find that -log(2.95 x 10^(-5)) ≈ 4.53. Plugging this value into the equation:
pH ≈ 14 - 4.53 = 9.47
Therefore, the pH of the solution with an [OH] of 2.95 x 10^-5 M is approximately 9.47.
3) To determine the pH at the equivalence point during a titration, we need to consider the stoichiometry of the reaction. In this case, acetic acid (CH3COOH) is being titrated with sodium hydroxide (NaOH).
The balanced chemical equation for the reaction is:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, we can see that one mole of acetic acid reacts with one mole of sodium hydroxide to form one mole of sodium acetate and one mole of water.
Since the volumes of both the acetic acid and NaOH used in the titration are given (25.00 mL each), we can determine the number of moles of acetic acid used.
moles of CH3COOH = volume of CH3COOH (L) * concentration of CH3COOH (M)
moles of CH3COOH = 0.025 L * 0.1056 M = 0.00264 moles
Since the reaction is a 1:1 ratio between acetic acid and sodium hydroxide, the moles of sodium hydroxide used will also be the same.
moles of NaOH = 0.00264 moles
To find the concentration of OH- at the equivalence point, we divide the moles of NaOH used by the total volume of the solutions:
NaOH concentration at equivalence point = moles of NaOH / total volume of solution (L)
total volume of solution = volume of CH3COOH (L) + volume of NaOH (L)
total volume of solution = 0.025 L + 0.025 L = 0.05 L
NaOH concentration at equivalence point = 0.00264 moles / 0.05 L = 0.0528 M
To determine the pH, we can use the equation:
pOH = -log([OH-])
pOH = -log(0.0528) = 1.28
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH = 14 - 1.28 ≈ 12.72
Therefore, the pH at the equivalence point when 25.00 mL of 0.1056 M acetic acid is titrated with 0.1056 M NaOH is approximately 12.72.
4) To determine the pH of a solution after a titration, we need to consider the stoichiometry and the volume of the solutions used. In this case, hydroiodic acid (HI) is being titrated with sodium hydroxide (NaOH).
The balanced chemical equation for the reaction is:
HI + NaOH -> NaI + H2O
From the equation, we can see that one mole of HI reacts with one mole of NaOH to form one mole of sodium iodide and one mole of water.
Given the volumes of HI and NaOH used in the titration (20.00 mL HI and 12.64 mL NaOH), we can determine the number of moles of HI and NaOH used.
moles of HI = volume of HI (L) * concentration of HI (M)
moles of HI = 0.020 L * 0.4963 M = 0.009926 moles
moles of NaOH = volume of NaOH (L) * concentration of NaOH (M)
moles of NaOH = 0.01264 L * 0.5174 M = 0.006546 moles
Since the reaction is a 1:1 ratio between HI and NaOH, the limiting reagent determines the number of moles of the product formed. In this case, NaOH is the limiting reagent because it has fewer moles.
From the balanced equation, we can see that when HI reacts with NaOH, it forms H2O. Since the volume of water formed is negligible compared to the volumes of the solutions used, the pH after the titration will depend on the excess of the remaining solution, which is NaOH.
To find the moles of OH- remaining, we subtract the moles of NaOH used from the initial moles of NaOH:
moles of OH- remaining = moles of NaOH - moles of NaOH used
moles of OH- remaining = 0.006546 moles - 0.006546 moles = 0 moles
Since there are no moles of OH- remaining, the concentration of OH- in the solution is 0 M.
To determine the pH, we can use the equation:
pH = 14 - pOH
Since the concentration of OH- is 0 M, pOH is also 0. Therefore, pH = 14 - 0 = 14.
Therefore, the pH of the solution after 20.00 mL of 0.4963 M HI has been titrated with 12.64 mL of 0.5174 M NaOH is 14.