Question

Calculate the ph of a buffer that is 0.225 m hc2h3o2 and 0.162 m kc2h3o2. the ka for hc2h3o2 is 1.8 Ã 10-5. 4.60 9.26 4.74 4.89 9.11

Answer 1

Answer:

The pH of the buffer solution is 4.60.

Explanation:

Concentration of acid = $[HC_2H_3O_2]=0.225 M$

Concentration of salt = $[KC_2H_3O_2]=0.162 M$

Dissociation constant = $K_a=1.8 \times 10^{-5}$

The pH of the buffer can be determined by Henderson-Hasselbalch equation:

$pH=pK_a+\log\frac{[salt]}{[acid]}$

$pH=-\log[1.8 \times 10^{-5}]+\log\frac{0.162 M}{0.225 M}$

pH = 4.60

The pH of the buffer solution is 4.60.

Answer 2

4.602

Further explanation

Given:

A buffer system consisting of 0.225 M HC₂H₃O₂ and 0.162 M KC₂H₃O₂.

The Ka for HC₂H₃O₂ is 1.8 x 10⁻⁵.

Question:

Calculate the pH of this buffer.

The Process:

Let us first observe the ionization reaction of the KC₂H₃O₂ salt below.

$\boxed{ \ KC_2H_3O_2 \rightleftharpoons K^+ + C_2H_3O_2^- \ }$

• The KC₂H₃O₂ salt has valence = 1 according to the number of C₂H₃O₂⁻ ions as a weak part.
• HC₂H₃O₂ and C₂H₃O₂⁻ are conjugate acid-base pairs
• HC₂H₃O₂ and C₂H₃O₂⁻ form an acidic buffer system.

To calculate the specific pH of a given buffer, we need using The Henderson-Hasselbalch equation for acidic buffers:

$\boxed{ \ pH = pK_a + log\frac{[A^-]}{[HA]} \ }$

where,  

• Ka represents the dissociation constant for the weak acid;
• [A-] represent the concentration of the conjugate base (i.e. salt);  
• [HA] is the concentration of the weak acid.

$\boxed{ \ pH = pK_a + log\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]} \ }$

$\boxed{ \ pH = -log(1.8 \times 10^{-5}) + log\frac{[0.162]}{[0.225]} \ }$

$\boxed{ \ pH = 5-log \ 1.8 - 0.1427 \ }$

$\boxed{ \ pH = 5 - 0.2553 - 0.1427 \ }$

$\boxed{ \ pH = 4.602 \ }$

Thus, the pH of this buffer equal to 4.602.

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