What is the pH of a solution of 120 ml 0.15M acetic acid to which we add 30mL 0.2M NaOH?
I have a bit of a hard time with this question. I tried to get mols of each one and then use it in an ice table but didn't work.
Can someone help? Thanks
Ice table ? I haven't heard of that term in awhile.
Well since acetic acid is a weak acid and NaOH is a strong base...
120ml 0.15M acetic acid
30ml 0.2M NaOH
-You need Ka for the weak acid.
(for acetic acid Ka= 1.8x10^-5)
steps:
1. Determine if moles H+ or moles OH- is in excess
2. Determine ammount of excess.
3. Determine Molarity of excess ammount.
4. Solve for pH directly or by pOH id OH- is in excess
Write the equation.
Calculate mols acetic acid.
Calculate mols NaOH.
Subtract them and see which is in excess. One of them will be left over. Probably,(my guess) acetic acid is left over and all the NaOH is added. Then you know how much sodium acetate is formed and this becomes a buffered solution. Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log[(base)/(acid)]
Post your work if you get stuck and I can help you through it.
Was what I posted incorrect Dr.Bob?
Step 1 was ok.
Step 2 was ok.
Step 3 is not correct because it forms a buffered solution and that can't be solved for pH directly; i.e., pH = -log(H^+) or pOH = -log(OH^-). The HH equation will do it OR the original Ka equation will solve for H^+ which can then be converted to pH.
Answered above. The first two steps are ok. The third is not.
Um actually it says this in my gen chem handout...not using the HH equation.
after determining the M of excess amount:
H+ + OH- => H2O
shifts right b/c product is removed.
using the numbers given in mine ..
-0.02500mol of HAc are consumed
-0.02500mol of Ac- is produced
-0.05000 - 0.02500= 0.02500moles HAc remain
then it says:
since this is a buffer (excess HAc[in this problem] and the NaOH caused it's conjugate base Ac- to be produced) we can use the initial moles (reduced aproximately) as the equillibrium moles
[H+]= Ka* moles HAc/moles Ac-= 1.8x10^-5
I just posted the general steps without elaborating but it was under determining the moles of the excess ammount but I assuem this is the original Ka equation?
1. Determine if moles H+ or moles OH- is in excess
2. Determine ammount of excess.
3. Determine Molarity of excess ammount.
4. Solve for pH directly or by pOH id OH- is in excess
I copied your original post above. I won't argue with you Christina for only you know what you meant; however,that isn't what you wrote. The implication in step 3 is that you have molarity of the acid or the molarity of the base and pH or pOH = -log(M) where M is either acid or base. It is also true that solving for (H^+) from the Ka expression gives (H^+) = Ka(acetic acid)/(acetate ion) and converting to pH will give the same answer. But that is where the H-H equation comes from and it gets the answer a little faster. Either equation is good.
Alright alright..I should be more specific next time or just let you answer it since I don't seem to be conveying it correctly. (I just copied what I had but I feel that the HH eqzn is easier to use than what I was first taught (this))
Thus the HH eqzn IS easier.
Of course, I'd be happy to help you with this question!
To determine the pH of the solution, we need to consider the reaction that occurs when acetic acid (CH3COOH) reacts with sodium hydroxide (NaOH). The reaction can be represented as follows:
CH3COOH + NaOH -> CH3COONa + H2O
In this reaction, acetic acid (CH3COOH) reacts with sodium hydroxide (NaOH) to form sodium acetate (CH3COONa) and water (H2O).
Since we know the initial concentrations and volumes of the acetic acid and sodium hydroxide, we can calculate the amount of each reactant that will react and determine the resulting concentrations of the products.
First, let's calculate the moles of acetic acid (CH3COOH) and sodium hydroxide (NaOH) present.
Moles of acetic acid (CH3COOH):
0.15 M x 0.120 L = 0.018 moles
Moles of sodium hydroxide (NaOH):
0.2 M x 0.030 L = 0.006 moles
Next, let's determine the limiting reactant. The reactant that is completely consumed is known as the limiting reactant. To find the limiting reactant, we compare the moles of acetic acid to sodium hydroxide. The ratio of acetic acid to sodium hydroxide in the balanced chemical equation is 1:1.
Since the reaction requires an equal number of moles of acetic acid and sodium hydroxide, and we have more moles of acetic acid (0.018 moles) than sodium hydroxide (0.006 moles), sodium hydroxide is the limiting reactant.
Now, let's calculate the moles of acetic acid and sodium hydroxide that reacted based on the stoichiometry of the balanced chemical equation.
Moles of acetic acid (CH3COOH) reacted:
0.006 moles (since sodium hydroxide is limiting)
Moles of sodium hydroxide (NaOH) reacted:
0.006 moles (stoichiometrically equal to the moles of acetic acid)
Since 0.006 moles of acetic acid reacted, and the final volume of the solution is 150 mL (120 mL + 30 mL), we can calculate the final concentration of acetic acid and sodium acetate.
Concentration of acetic acid (CH3COOH) after reaction:
0.006 moles / 0.150 L = 0.04 M
Concentration of sodium acetate (CH3COONa) after reaction (since the stoichiometry is 1:1):
0.006 moles / 0.150 L = 0.04 M
To determine the pH of the solution, we need to consider the dissociation of acetic acid (CH3COOH) into its conjugate base, acetate (CH3COO-), and a hydrogen ion (H+). Acetic acid is a weak acid, so it only partially dissociates in water.
The dissociation of acetic acid can be represented as follows:
CH3COOH ⇌ CH3COO- + H+
The equilibrium constant expression for this reaction is given by the equation:
Ka = [CH3COO-][H+]/[CH3COOH]
In this equation, [CH3COO-] represents the concentration of acetate, [H+] represents the concentration of hydrogen ions, and [CH3COOH] represents the concentration of acetic acid.
The Ka value for acetic acid is approximately 1.8 x 10^-5 at 25°C.
Since the concentration of acetic acid after the reaction is 0.04 M, and we assume the concentration of acetate and hydrogen ions at equilibrium is x M, we can set up the equilibrium expression as follows:
1.8 x 10^-5 = x^2 / 0.04
Simplifying the equation, we have:
x^2 = (1.8 x 10^-5) x 0.04
x^2 = 7.2 x 10^-7
Taking the square root of both sides, we find:
x = 2.7 x 10^-4 M
Since the concentration of hydrogen ions is equal to the concentration of acetate after the reaction, the pH of the solution can be determined using the equation:
pH = -log[H+]
pH = -log(2.7 x 10^-4)
pH ≈ 3.57
Therefore, the pH of the solution is approximately 3.57.