Answer:
The correct answer is 4.84
Explanation:
We use the Henderson Hasselbach's equation to calculate the pH of a buffer solution:
pH = pKa = log ( [conjugate base]/[weak acid])
In this case:
conjugate base: acetate ⇒ [conjugate base] = [acetate] = 0.0200 M
weak acid: acetic acid ⇒ [weak acid] = [acetic acid] = 0,0200 M
pH = pKa + log ([acetate]/[acetic acid])= pKa + log (0.0200 M/0.0200 M)
When a strong base- such as NaOH- is added, the acid reacts with OH⁻ to form the conjugate base. So, the conjugate base is increased while the acid is decreased. Thus, in 1 liter of solution we have:
acetate = (0,0200 mol/L x 1 L) + (2.0 mmol x 1 mol/1000 mmol) = 0.022 mol
acetic acid= (0,0200 mol/L x 1 L) - (2.0 mmol x 1 mol/1000 mmol) = 0.018 mol
Finally, we calculate the pH:
pH = 4.75 + log (0.022 mol /0.018 mol )= 4.84
Buffer has been the solution that resists the change in the pH of the solution with the addition of acid or base. The pH of buffer with NaOH has been 4.84.
The pH has been the hydrogen ion concentration in the solution. The pH has been equivalent to pKa and can be given as:
[tex]\rm pka=log\dfrac{[base]}{[acid]}[/tex]
The base in the buffer has been the acetate, while acetic acid act as the acid.
The concentration of NaOH added to the solution has been given as:
[tex]\rm Molarity=\dfrac{moles}{volume}\\\\ NaOH=\dfrac{0.002\;mol}{1\;L}\\\\ NaOH=0.002\;M[/tex]
The addition of NaOH adds the concentration of base, and neutralization reduces the concentration of acid. Thus, the new pKa can be given as:
[tex]\rm pKa=log\dfrac{0.2+0.002}{0.2-0.002} \\pKa=log\dfrac{0.022}{0.018}[/tex]
The new pH of the solution has been the sum of the old and the new pKa value.
Thus, the pH of the new buffer has been given as:
[tex]\rm pH=4.75+log\dfrac{0.022}{0.018}\\ pH=4.84[/tex]
The pH of buffer with NaOH has been 4.84.
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