Respuesta :
When 0.195 mole of NaOH are added and there is no volume change, moles of base will increase by 0.195 and moles of acid will decrease by 0.195 mole
[Base] = 0.608 + 0.195 = 0.803 mole
[Acid] = 0.809 - 0.195 = 0.614 mole
pH = 6.25 + log [tex] \frac{0.803}{0.614} = 6.37[/tex]
[Base] = 0.608 + 0.195 = 0.803 mole
[Acid] = 0.809 - 0.195 = 0.614 mole
pH = 6.25 + log [tex] \frac{0.803}{0.614} = 6.37[/tex]
pH after addition of NaOH = 6.364
Further explanation
A buffer solution is a solution that can maintain a good pH value due to the addition of a little acid or a little base or dilution.
The buffer solution can be acidic or basic
• Acid buffer solutions consist of weak acids and their salts.
[tex]\displaystyle [H ^ +] = Ka \times \frac {mole \: weak \: acid} {mole \: salt \times valence}[/tex]
valence according to the amount of salt anion
• Alkaline buffer solution consists of a weak base and its salt.
[tex]\displaystyle [OH ^ -] = Kb \times \frac {mole \: weak \: base} {mole \: salt \times valence}[/tex]
valence according to the amount of salt cation
• If the buffer solution is acidic then
a slight addition of acid (H⁺) will be balanced by the conjugate base
the addition of a small base (OH⁻) will be balanced by the weak acid
• If the buffer solution is alkaline then
the addition of a little acid (H⁺) will be balanced weak base
the addition of a small base (OH⁻) will be balanced
the weak acid HA and NaA, including Acid buffer solution
the addition of a small base (OH⁻) will be balanced by the weak acid
Part A: What is the pH of a buffer prepared by adding 0.809 moles of the weak acid HA to 0.609 moles of NaAin 2.00 L of solution? The dissociation constant Ka of HA is 5.66×10−7
The ph after 0.195 mol of naoh is added to the buffer from part A:
Reactions that occur:
HA + NaOH ---> NaA + H2O
0.809 0.195 0.609
0.195 0.195 0.195
0.614 0 0.804
[tex]\displaystyle [H^+]=Ka \times \frac {mole\: weak\:acid}{mole \:salt \times valence}[/tex]
[tex]\displaystyle [H^+]=5.66.10^{-7} \times \frac {0.614}{0.804\times 1}[/tex]
[tex]\displaystyle pH=-log~4.322.10^{-7}\\\\pH=6.364[/tex]
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Keywords : buffer, acid, salt, pH