acetic acid: 0.1 M 30ml
hydrochloric acid 0.1 M 5 ml
sodium hydroxide 0.1 M 15 ml
Buffer + 5ml of: 1.91 ph
ph (measured): 4:52 ph
HOW DO I CALCULATE PH(CALCULATED)?
AND ALSO IT SAYS
SHOW YOUR CALCULATION FOR THE PH OF THE BUFFER BEFIRE AND AFTER THE ADDITION OF HCL?
COULD YOU PLEASE HELP ME THANK YOU!
I could help but your post is too disjointed and some information seems to be missing. Please repost the problem as it appears in your text. If this is a lab experiment, show how the buffer is made, what it consists of, etc.
Okay let me write it this way.
I mixed 30 mL of 0.1 M Acetic Acid and 15 mL of 0.1 M Sodium Hydroxide and measured the pH. Then I added 5 mL of 0.1 M NaOH to this buffer and measured the pH.
It's asking me:
Show the calculations for the pH of the buffer before and after the addition of HCL?
Thank You
OK. Sorry for the delay in getting back to you but I had a concert to attend.
30 mL x 0.1M HAc = 3 millimoles.
15 mL x 0.1M NaOH = 1.5 millimols.
......NaOH + HAc ==> NaAc + H2O
I.....1.5......3.0....0.......0
C.....-1.5.....-1.5...1.5......0
E.......0......1.5.....1.5
Thus, the buffer consists of 1.5 mmols Ac^- and 1.5 mmols HAc. The pH of this buffer is pH = pKa + log(base)/(acid) = pKa + log(1.5/1.5) = pKa + log 1 = pKa + 0 = pKa = 4.74 approximately but you need to look up Ka in your tables and use that value. I used 1.8E-5 to obtain 4.74.
If we add 5 mL of 0.1M HCl we add 0.5 mmols H^+.
..........Ac^- + H^+ ==> HAc
I........1.5......0.......1.5
add ..............0.5..........
C.......-0.5.....-0.5.....+0.5
E.......1.0........0......2.0
Then recalculate the pH with the H-H equation.
pH = pKa + log(1/2) = ?
I how you answers this but what is "c" HAc? NaAc?
To calculate the pH (calculated) of a buffer solution, you need to consider the initial concentrations of the acid and its conjugate base in the buffer solution and the equilibrium constant of their dissociation reaction.
For the calculation, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH is the measure of hydrogen ion concentration,
pKa is the negative logarithm of the acid dissociation constant (Ka) of the acid component of the buffer,
[A-] is the molar concentration of the conjugate base,
[HA] is the molar concentration of the acid.
In your case, acetic acid (CH3COOH) and sodium acetate (CH3COONa) are involved in the buffer solution. Acetic acid is a weak acid, and its pKa is around 4.75.
To calculate the pH (calculated), we need to determine the concentrations [A-] and [HA]:
[A-] = moles of sodium acetate / total volume of the buffer solution
= 0.1 M * 15 mL / (30 mL + 5 mL + 15 mL)
= 0.1 M * 15 mL / 50 mL
= 0.03 M
[HA] = moles of acetic acid / total volume of the buffer solution
= 0.1 M * 30 mL / (30 mL + 5 mL + 15 mL)
= 0.1 M * 30 mL / 50 mL
= 0.06 M
Now we can substitute these values into the Henderson-Hasselbalch equation:
pH = 4.75 + log(0.03 / 0.06)
= 4.75 + log(0.5)
= 4.75 - 0.301
= 4.449
So, the calculated pH of the buffer before the addition of any other substance is approximately 4.449.
To calculate the pH of the buffer after the addition of hydrochloric acid (HCl), you need to consider the reaction between the HCl and the acetate ion.
HCl is a strong acid, so it will fully dissociate in water, leading to an increase in the concentration of hydrogen ions (H+) in the solution. However, since the buffer consists of acetic acid (a weak acid) and its conjugate base, the buffer solution can resist changes in pH to some extent.
The reaction between HCl and the acetate ion can be written as:
HCl + CH3COO- -> CH3COOH + Cl-
The added HCl reacts with the acetate ion to form acetic acid and chloride ions. The concentration of acetic acid increases, and the concentration of the acetate ion decreases.
To calculate the pH after the addition of HCl, you need to know the change in concentration of acetic acid ([HA]) and the change in concentration of the acetate ion ([A-]).
Since 5 mL of a 0.1 M HCl solution is added to a total volume of 50 mL, the final concentration of HCl ([HCl]) will be:
[HCl] = (0.1 M * 5 mL) / 50 mL
= 0.01 M
The change in concentration of acetic acid ([HA]) will be:
Change in [HA] = [HCl]
The change in concentration of the acetate ion ([A-]) will be equal to the change in [HA] but with the opposite sign:
Change in [A-] = -[HCl]
Now, we can update the concentrations of acetic acid ([HA]) and the acetate ion ([A-]) in the Henderson-Hasselbalch equation and calculate the pH:
pH = 4.75 + log(([A-] - [HCl]) / ([HA] + [HCl]))
= 4.75 + log(([0.03 M] - [0.01 M]) / ([0.06 M] + [0.01 M]))
= 4.75 + log(0.02 M / 0.07 M)
≈ 4.75 + log(0.2857)
≈ 4.75 - 0.545
= 4.205
Therefore, the calculated pH of the buffer after the addition of the HCl solution is approximately 4.205.