Respuesta :
To calculate the pH of this solution, we use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where,
[A-] = Molarity of the conjugate base =
CH3COO- = 0.29 M
[HA] = Molarity of the weak acid = CH3COOH = 0.18 M
pKa = dissociation constant of the weak acid = 4.75
When KOH is added to the buffer, the chemical reaction is:
CH3COOH + KOH = CH3COO-K+ + H2O
Therefore when 0.0090 mol KOH is added, 0.0090 mol acid is neutralized, and 0.0090 mol CH3COO- is produced.
[CH3COO-] = [0.0090 mol + 0.375 L (0.29 mol/L) ] / 0.375 L = 0.314 M
[CH3COOH] = [-0.0090 mol + 0.375 L (0.18 mol/L) ] / 0.375 L = 0.156 M
Going back to Henderson-Hasselbalch equation:
pH = 4.75 + log (0.314 / 0.156)
pH = 5.054
The pH of 0.375 L of a 0.18 M acetic acid-0.29 M sodium acetate buffer is [tex]\boxed{5.054}[/tex].
Further Explanation:
Buffer solution:
It is an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.Any change in the pH of such solutions is opposed when small quantities of strong acid or base are added to them.
Henderson-Hasselbalch equation:
This is used for the determination of pH of buffer solution. Its mathematical form is given as follows:
[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}[/tex] …… (1)
Here,
[tex]\left[ {{{\text{A}}^ - }} \right][/tex] is the concentration of conjugate base.
[HA] is the concentration of acid.
The given mixture is a buffer solution of acetic acid and sodium acetate. Therefore Henderson-Hasselbalch equation becomes,
[tex]{\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }} \right]}}{{\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}} \right]}}[/tex] …… (2)
Initial moles of [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }[/tex] can be calculated as follows:
[tex]\begin{aligned} {\text{Moles of C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } &= \left( {0.29{\text{ M}}} \right)\left( {{\text{0}}{\text{.375 L}}} \right) \\ &= 0.10875{\text{ mol}} \\ \end{aligned}[/tex]
Initial moles of [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}[/tex] can be calculated as follows:
[tex]\begin{aligned} {\text{Moles of C}}{{\text{H}}_{\text{3}}}{\text{COOH}} &= \left( {0.18{\text{ M}}} \right)\left( {{\text{0}}{\text{.375 L}}} \right) \\ &= 0.0675{\text{ mol}} \\ \end{aligned}[/tex]
When 0.0090 moles of KOH are added to the buffer solution, 0.0090 moles of acetic acid is neutralized while the same amount of acetate ions are formed.
Therefore concentration of [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }[/tex] can be calculated as follows:
[tex]\begin{aligned} \left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }} \right] &= \frac{{\left( {0.10875 + 0.0090} \right){\text{ mol}}}}{{0.375{\text{ L}}}} \\ &= 0.314{\text{ M}} \\ \end{aligned}[/tex]
Therefore concentration of [tex]{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}[/tex] can be calculated as follows:
[tex]\begin{aligned} \left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}} \right] &= \frac{{\left( {0.0675 - 0.0090} \right){\text{ mol}}}}{{0.375{\text{ L}}}} \\ &= 0.156{\text{ M}} \\ \end{aligned}[/tex]
Substitute 0.156 M for [tex]\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}} \right][/tex] , 0.314 M for [tex]\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }} \right][/tex] and 4.75 for [tex]{\text{p}}{K_{\text{a}}}[/tex] in equation (2).
[tex]\begin{aligned} {\text{pH}} &= 4.75 + {\text{log}}\left( {\frac{{0.314{\text{ M}}}}{{0.156{\text{ M}}}}} \right) \\ & = 5.054 \\ \end{aligned}[/tex]
Learn more:
- The reason for the acidity of water https://brainly.com/question/1550328
- Reason for the acidic and basic nature of amino acid. https://brainly.com/question/5050077
Answer details:
Grade: High School
Chapter: Acid, base and salts.
Subject: Chemistry
Keywords: pH, buffer, acetate, acetic acid, KOH, 5.054, pKa, 0.156 M, 4.75, 0.314 M.
