1L of a buffer composed of acetic acid and sodium acetate has a pH of 4.3. Adding 10mL of 2M sodium hydroxide solution to 100mL of this buffer causes the pH to rise to 4.87. what is the total molarity of the original buffer?
pH = pka + log b/a
4.3 = 4.74 + log b/a
b/a = 0.363
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add 100 mL x 2M = 20 mmols to obtain
4.87 = 4.74 + log b/a
b/a = 1.35 and redo to read
1.349 = (b+20)/(a-20)
1.349(a-20) = b+20
1.349a - 27 = b+20
Substitute 0.363 for b to obtain
1.349a - 27 = 0.363a + 20
0.986a = 47
a = 47.7
b = 0.363a = 0.363*47.7 = 17.3
a + b = 47.7 + 17.3 = 65
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Check:
..........HAc + OH^- ==> Ac^- + H2O
I.........47.7...0.......17.3
add ............20.........
C.........-20..-20.......+20
E.........27.7...0.......37.3
pH = 4.74 + log(37.3/27.7)
pH = 4.87
i got a total concentration of 0.0774M is this correct?
Where are you getting that?
To find the total molarity of the original buffer, we need to use the Henderson-Hasselbalch equation. The equation is given as:
pH = pKa + log([A-]/[HA])
Where:
pH is the pH of the buffer,
pKa is the dissociation constant of the acid, and
[A-] and [HA] are the concentrations of the conjugate base and acid, respectively.
Given that the pH of the buffer solution is 4.3, we can use this value to find the pKa of acetic acid, as acetic acid is the acid component of the buffer.
pH = pKa + log([A-]/[HA])
4.3 = pKa + log([A-]/[HA])
Since we have a buffer solution, the concentrations of the conjugate base ([A-]) and acid ([HA]) are in a 1:1 ratio. So, [A-] = [HA].
4.3 = pKa + log(1)
We can rearrange the equation to solve for pKa:
pKa = 4.3
Now, let's calculate the change in pH that occurs when 10 mL of 2M sodium hydroxide (NaOH) solution is added to 100 mL of the buffer. The initial pH before adding NaOH is 4.3 and the final pH after adding NaOH is 4.87. Therefore:
Change in pH = final pH - initial pH
Change in pH = 4.87 - 4.3
Change in pH = 0.57
Since the change in pH is positive, it indicates a shift towards the basic side. This means that the concentration of the conjugate base ([A-]) has increased, and the concentration of the acid ([HA]) has decreased.
We know the volume of the buffer solution (1L), so the moles of NaOH added can be calculated using the formula:
Moles = concentration (M) x volume (L)
Moles of NaOH added = 2 M x 0.01 L
Moles of NaOH added = 0.02 mol
Since NaOH reacts in a 1:1 ratio with acetic acid (HA) to form sodium acetate (A-), the moles of acetic acid (HA) used up can be calculated as well. Therefore, the moles of sodium acetate (A-) formed can also be calculated.
Moles of HA used up = 0.02 mol
Moles of A- formed = 0.02 mol
Now we need to determine the initial concentrations of [HA] and [A-] before adding NaOH. Since we know the moles and volume of the buffer solution, we can calculate the molarities:
Molarity = moles / volume (L)
Molarity of HA = moles of HA / volume of buffer solution
Molarity of A- = moles of A- / volume of buffer solution
Molarity of HA = 0.02 mol / 1 L
Molarity of A- = 0.02 mol / 1 L
Both the molarities of [HA] and [A-] equal 0.02 M.
Therefore, the total molarity of the original buffer is the sum of the molarities of [HA] and [A-]:
Total molarity of the original buffer = Molarity of HA + Molarity of A-
Total molarity of the original buffer = 0.02 M + 0.02 M
Total molarity of the original buffer = 0.04 M
So, the total molarity of the original buffer is 0.04 M.