Question

Calculate the ph of 0.375 l of a 0.18 m acetic acid-0.29 m sodium acetate buffer after the addition of 0.0090 mol of koh. assume that the volume remains constant.

Answer 1

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

Answer 2

The pH of 0.375 L of a 0.18 M acetic acid-0.29 M sodium acetate buffer is $\boxed{5.054}$.

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:

${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}$                                            …… (1)

Here,

$\left[ {{{\text{A}}^ - }} \right]$ 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,

${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }} \right]}}{{\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}} \right]}}$                                                  …… (2)

Initial moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }$ can be calculated as follows:

$\begin{aligned} {\text{Moles of C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } &= \left( {0.29{\text{ M}}} \right)\left( {{\text{0}}{\text{.375 L}}} \right) \\ &= 0.10875{\text{ mol}} \\ \end{aligned}$  

Initial moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$  can be calculated as follows:

$\begin{aligned} {\text{Moles of C}}{{\text{H}}_{\text{3}}}{\text{COOH}} &= \left( {0.18{\text{ M}}} \right)\left( {{\text{0}}{\text{.375 L}}} \right) \\ &= 0.0675{\text{ mol}} \\ \end{aligned}$  

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 ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }$ can be calculated as follows:

$\begin{aligned} \left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }} \right] &= \frac{{\left( {0.10875 + 0.0090} \right){\text{ mol}}}}{{0.375{\text{ L}}}} \\ &= 0.314{\text{ M}} \\ \end{aligned}$  

Therefore concentration of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ can be calculated as follows:

$\begin{aligned} \left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}} \right] &= \frac{{\left( {0.0675 - 0.0090} \right){\text{ mol}}}}{{0.375{\text{ L}}}} \\ &= 0.156{\text{ M}} \\ \end{aligned}$  

Substitute 0.156 M for $\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}} \right]$ , 0.314 M for $\left[ {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }} \right]$ and 4.75 for ${\text{p}}{K_{\text{a}}}$ in equation (2).

$\begin{aligned} {\text{pH}} &= 4.75 + {\text{log}}\left( {\frac{{0.314{\text{ M}}}}{{0.156{\text{ M}}}}} \right) \\ & = 5.054 \\ \end{aligned}$  

Learn more:

1. The reason for the acidity of water brainly.com/question/1550328
2. Reason for the acidic and basic nature of amino acid. 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.