how many mL of .100 M of acetic acid and .100 M of NaOH, would be need to make the volume 1.0 liters and have a pH7.00 if Ka=1.8x10^-5
please write out the steps or eq for me im gettin stuck
This is not a good question because the pH of 7.00 is too far from the pKa value to work well as a buffer. However, here is the problem.
pH = pKa + log (base)/(acid)
7.00 = 4.74 + log (B/A)
Solve for (base/(acid). I obtained 181.97 but you need to confirm that. Here base is referring to acetate ion (not NaOH) and acid is referring to acetic acid. I call those HAc for acetic acid and Ac^- for acetate (the base).
Then you set up another equation.
moles base = 181.97*moles acid
For moles base substitute mLbase x 0.1 M. On the right side, substitute for moles acid (1000 - mLbase)*0.1 M
Solve for mLbase
I get something like 994.53 for base and 1000-994.53 = 5.47 mL acid. You need to confirm both numbers AND you may need to carry out to more places that you are allowed under significant figure rules BECAUSE it is such a large number for one and a small number for the other one.
What this means is that you need to add that many mL of 0.1 M NaOH to produce that much Ac^- and since it is neutralizing HAc, the HAc is being used up at the same time. To check if this is correct,
(Ac^-) formed = 994.53 mL x 0.1M/1000 mL = 0.099453 M.
(HAc) remaining = 5.465 mL x 0.1 M/1000 = 0.0005465
pH = 4.74 + log (0.099453/0.0005465) =
pH = 4.74 + 2.26 = 7.00. I reiterate this is a poor problem AND that I have used more significant figures than allowed. With such a disparity in numbers, that is necessary to make the numbers come out right.
I don't think this would be correct because The solution you are proposing would have a pH over 12.
To find the volume of acetic acid and NaOH required to make a solution with a pH of 7.00, we first need to understand the chemistry behind it.
Acetic acid (CH3COOH) is a weak acid, and when it reacts with water, it dissociates into its ionized form as follows:
CH3COOH + H2O ⇌ CH3COO- + H3O+
Similarly, NaOH is a strong base that completely dissociates into its ionized form in water:
NaOH ⇌ Na+ + OH-
To achieve a solution with a pH of 7.00, we need to neutralize the hydrogen ions (H+) present in the solution. This can be done by reacting the acetic acid (which donates H+) with an equivalent amount of NaOH (which accepts H+) to form water.
The balanced chemical equation for the reaction between acetic acid and NaOH is:
CH3COOH + NaOH → CH3COONa + H2O
Now, let's calculate the moles of acetic acid required using the equation:
moles = Molarity × Volume (in liters)
Since the final volume is 1.0 liter, we need to determine the moles of acetic acid and NaOH based on the balanced equation and their respective molarities.
Step 1: Calculate the moles of acetic acid required
moles of acetic acid = (required OH- ions) = (required H+ ions)
The concentration of OH- ions in a solution is equal to the concentration of H+ ions required to neutralize them to achieve pH 7.00. Hence, we need to calculate the concentration of H+ ions.
The concentration of H+ ions can be determined using the equation for the dissociation of acetic acid, assuming it reaches equilibrium:
[H+] × [CH3COO-] / [CH3COOH] = Ka
Given that Ka (acid dissociation constant) = 1.8 × 10^-5, we can rearrange the equation to solve for [H+]:
[H+]^2 / [CH3COOH] = Ka
[H+]^2 / [CH3COOH] = 1.8 × 10^-5
[H+]^2 = 1.8 × 10^-5 × [CH3COOH]
[H+] = √(1.8 × 10^-5 × [CH3COOH])
Since [CH3COOH] is 0.100 M (given in the problem), we can substitute it into the equation to find [H+]:
[H+] = √(1.8 × 10^-5 × 0.100)
[H+] = √(1.8 × 10^-6) = 4.24 × 10^-4 M
Now that we know the concentration of H+ ions, we can calculate the moles of acetic acid required:
moles of acetic acid = 4.24 × 10^-4 M × 1.0 L = 4.24 × 10^-4 mol
Step 2: Calculate the moles of NaOH required
Since the balanced equation shows a 1:1 stoichiometric ratio between acetic acid and NaOH, the moles of NaOH required will also be 4.24 × 10^-4 mol.
Finally, to calculate the volume of each solution required, we can use the equation:
Volume (in mL) = (moles × 1000) / Molarity
Volume of acetic acid = (4.24 × 10^-4 mol × 1000 mL) / 0.100 M = 4.24 mL
Volume of NaOH = (4.24 × 10^-4 mol × 1000 mL) / 0.100 M = 4.24 mL
Therefore, to make a 1.0 liter solution with a pH of 7.00, you would need to mix 4.24 mL of 0.100 M acetic acid and 4.24 mL of 0.100 M NaOH.