Question

Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq)

Answer 1

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

$\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100$    

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].          

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq)   (1)

0.1 - x                                         x                     x

$Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$  (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

$1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x}$  

$1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0$

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:      

$\% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \%$                      

   

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

$Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}$

With [CH₃COOH] = 0.01 M

$1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x}$  

$1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0$

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]    

Then, the percent ionization for this case is:

$\% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \%$

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!