Question

The ka of acetic acid (hc2h3o2) is 1.8 ⋅ 10-5. what is the ph at 25.0 °c of an aqueous solution that is 0.100 m in acetic acid? the ka of acetic acid (hc2h3o2) is 1.8 10-5. what is the ph at 25.0 °c of an aqueous solution that is 0.100 m in acetic acid? +2.87 -2.87 -11.13 +6.61 +11.13

Answer 1

Answer:

2.872.

Explanation:

• For weak acids: [H⁺] = √(Ka.C).

∴ [H⁺] = √(Ka.C) = √(1.8 x 10⁻⁵)(0.1 M) = 1.342 x 10⁻³.

∵ pH = - log[H⁺].

∴ pH = - log(1.342 x 10⁻³) = 2.872.

Answer 2

Answer : The pH of the solution is, 2.87

Solution :  Given,

Concentration (c) = 0.100 M

Acid dissociation constant = $k_a=1.8\times 10^{-5}$

The equilibrium reaction for dissociation of $CH_3COOH$ is,

                           $CH_3COOH\rightleftharpoons CH_3COO^-+H^+$

initially conc.         c                       0              0

At eqm.              $c(1-\alpha)$                $c\alpha$             $c\alpha$

First we have to calculate the concentration of value of dissociation constant $(\alpha)$.

Formula used :

$k_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}$

Now put all the given values in this formula ,we get the value of dissociation constant $(\alpha}$.

$1.8\times 10^{-5}=\frac{(0.1\alpha)(0.1\alpha)}{0.1(1-\alpha)}$

By solving the terms, we get

$\alpha=0.0133$

Now we have to calculate the concentration of hydrogen ion.

$[H^+]=c\alpha=0.1\times 0.0133=1.33\times 10^{-3}M$

Now we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.33\times 10^{-3})$

$pH=2.87$

Therefore, the pH of the solution is, 2.87