1)_What is the pH of the medium when 10.00 mL of 0.1000M NaOH is added to 18.00 mL of a 0.1000M benzoic acid (C6H5COOH) solution?
(Ka = 6.28×10^-5 for C6H5COOH)
2)_How many grams of NHCl (53.491 g/mol) should be added to 200 mL of 0.180M NH3 solution in order to prepare a buffer solution with pH 8.5?
Let's call benzoic acid HBz. Then HBz + NaOH --> NaBz + H2O and this is a buffer solution with NaBz as the base and HBz as the acid.
Millimols NaOH added = millimols NaBz formed = mL x M = 10 x 0.1 = 1
millimols HBz initially = 18 x 0.1 = 1.8
........................HBz + NaOH ==> NaBz + H2O
I.......................1.8.........0................0...............
added..........................0.1.......................................
C....................-0.1.....-0.1..............+0.1........................
E......................0.8.........0...............0.1.....................
So (HBz) @ equilibrium = millimoles/mL = 0.8/28 = ?
(NaBz) @ equilibrium = 0.1/28 = ?
Plug this into the HH equation as
pH = pKa for HBz + log [(NaBz)/(HBz)]
2. I assume you meant NH4Cl in that first line.
Kb for NH3 = 1.75E-5 but use the Ka in your text/notes. Then pKb = 4.76 and pKa + pKb = pKw = 14 so pKa = 9.24
pH = pKa + log [(NH3)/(NH4Cl)]
8.5 = 9.24 + log [(0.18)/(NH4Cl)]
-0.74 = log 0.18/(NH4Cl)
0.182 = 0.18/(NH4Cl)
(NH4Cl) = 0.18/0.182 = about 1 M for NH4Cl
1M NH4Cl will be about 54 g/1000 mL or 54/5 = about 10.8 g NH4Cl in 200 mL. I've estimated here and there but you can clean that up. NOTE, There are so many places to go wrong in a problem like this that I N EVER let it go without checking. So let's check it by placing 10.8 g NH4Cl in 200 mL of 0.18 M NH3 and see if it give a pH of 8.5.
10.8 g NH4Cl in 200 mL.
10.8/54 = 0.2 mol NH4Cl and that in 0.200 L = 1 M
pH = pKa + log b/a
pH = 9.24 + log 0.18/1
pH = 9.24 - 0.74 = 8.5 YEA!!!
Post your work if you get stuck.
thank you very much🙏🏻☺️
To find the pH of a solution or to prepare a buffer solution, we need to understand the concept of acid-base reactions and the relevant equations. Let's go through the steps to answer each question:
1) To determine the pH of the medium when NaOH is added to a benzoic acid solution, we need to consider the equilibrium reaction between the benzoic acid (C6H5COOH) and NaOH. The balanced equation for this reaction is:
C6H5COOH + NaOH ⇌ C6H5COONa + H2O
Since benzoic acid is a weak acid, it does not completely ionize in water. However, NaOH is a strong base that fully dissociates into its ions (Na+ and OH-) in water.
To determine the pH, we need to calculate the concentration of the hydronium ion (H3O+) in the final solution. The concentration can be obtained from the reaction's stoichiometry. In this case, since NaOH and benzoic acid have a 1:1 mole ratio in the balanced equation, the change in the concentration of benzoic acid will be equal to the change in the concentration of hydronium ions.
We start by calculating the initial moles of benzoic acid and NaOH:
moles of benzoic acid = volume (L) × concentration (M)
= 18.00 mL × 0.1000 M
= 0.00180 moles
moles of NaOH = volume (L) × concentration (M)
= 10.00 mL × 0.1000 M
= 0.00100 moles
Since NaOH is a strong base, it will react completely with benzoic acid. Thus, the number of moles of benzoic acid remaining after the reaction will be:
moles of benzoic acid remaining = moles of initial benzoic acid - moles of NaOH used
= 0.00180 moles - 0.00100 moles
= 0.00080 moles
To calculate the concentration of benzoic acid remaining after the reaction, divide the moles by the total volume of the solution (in liters):
concentration of benzoic acid remaining = moles of benzoic acid remaining / total volume (L)
= 0.00080 moles / (18.00 mL + 10.00 mL)
= 0.00080 moles / 0.02800 L
= 0.02857 M
Now, we can calculate the concentration of the hydronium ion (H3O+) in the final solution, which is equal to the concentration of benzoic acid remaining:
[H3O+] = 0.02857 M
To find the pH, take the negative base-10 logarithm of the hydronium ion concentration:
pH = -log[H3O+]
= -log(0.02857)
= 1.546
Therefore, the pH of the medium when 10.00 mL of 0.1000 M NaOH is added to 18.00 mL of a 0.1000 M benzoic acid solution is approximately 1.546.
2) To determine the mass of NH4Cl needed to prepare a buffer solution with pH 8.5, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa (the negative base-10 logarithm of the acid dissociation constant, Ka) and the ratio of the concentrations of the weak acid (NH4+) and its conjugate base (NH3) in the buffer solution.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([NH3] / [NH4+])
Rearranging the equation, we have:
log([NH3] / [NH4+]) = pH - pKa
First, we need to calculate the concentration of NH4+ in the buffer solution. It is equal to the concentration of NH3 because NH4Cl will dissociate into NH4+ and Cl- ions.
We can use the Henderson-Hasselbalch equation to find the ratio [NH3] / [NH4+] by substituting the given pH and the pKa value (which can be found using the given Ka value).
pH = 8.5
pKa = -log(Ka) = -log(6.28×10^-5) = 4.202
log([NH3] / [NH4+]) = 8.5 - 4.202
= 4.298
Now, we can take the antilog of both sides of the equation:
[NH3] / [NH4+] = 10^4.298
= 23344.10
Since the concentration of NH3 is already given (0.180 M), we can solve for the concentration of NH4+:
[NH4+] = [NH3] / (NH3 / NH4+)
= 0.180 M / 23344.10
= 7.709 × 10^-6 M
To convert the concentration (Molar) to moles, multiply by the volume of the solution:
moles of NH4+ = (7.709 × 10^-6 M) × 0.200 L
= 1.542 × 10^-6 mol
To calculate the mass of NH4Cl needed, we multiply the number of moles by its molar mass:
mass of NH4Cl = moles of NH4+ × molar mass of NH4Cl
= (1.542 × 10^-6 mol) × 53.491 g/mol
= 8.249 × 10^-5 g
Therefore, approximately 8.249 × 10^-5 grams of NH4Cl should be added to 200 mL of 0.180 M NH3 solution to prepare a buffer solution with pH 8.5.