Question

Calculate the percent ionization of a 0.15 M benzoic acid solution in pure water and in a solution containing 0.10 M sodium benzoate. Why does the percent ionization differ significantly in the two solutions

Answer 1

Answer:

% ionization for benzoic acid = 0.08%

% ionization for sodium benzoate = 2.5%

The percentage ionization differ significantly because benzoic acid is a weak acid while sodium benzoate is a salt of benzoic acid. Their extent of dissociation also differ because they were compared in different solutions

Explanation:

Ka for pure water = 1.0 * 10-⁷

Ka for sodium benzoate = 6.5*10-⁵

1. For benzoic acid (C6H5COOH)

C6H5COOH ==== C6H5COO‐ + H+

0.15M 0 0

0.15-x x x

Ka = [C6H5COO-] [H+] / [C6H5COOH]

Ka = [X] [X] / 0.15 - X

1.0*10-⁷ = [X]² / 0.15 - x

But x is negligible compared to 0.15,

(1.0*10-⁷)*0.15 = x²

Take square root of both sides,

X = 1.22 * 10-⁴

% ionization = ( [H+] / [C6H5COOH] ) * 100

% ionization = (1.22*10-⁷ / 0.15) * 100

% ionization = 0.08%

2. For C6H5COONa

Note: I will not repeat the same procedure of dissociation again since they're basically the same just the difference in ions

Ka for C6H5COONa = 6.5*10-⁵

6.5*10-⁵ = [X]² / (0.10 - X)

Cross-multiply both sides;

(6.5*10-⁵ * 0.10) = X²

Take square root of both side,

X= 2.5*10-³

% ionization = (2.5*10-³ / 0.10) *100

% ionization = 2.5%

Answer 2

Ionization is the gain or loss of an electron by the atom or the molecule. The percentage ionization for benzoic is 0.08% and for sodium benzoate is 2.5%.

What is Ionization?

Ionization is the ability of the atom or the molecule to form a cation or anion by losing or gaining electrons.

Given,

Ka for pure water = $1.0 \times 10^{-7}$

Ka for sodium benzoate = $6.5 \times 10^{-5}$

The percentage ionization for benzoic acid can be calculated as:

$\begin{aligned}\rm Ka &= \rm \dfrac{[C_{6}H_{5}COO^{-}] [H^{+}]}{[C_{6}H_{5}COOH]}\\\\1.0 \times 10^{-7} &= \rm \dfrac{[X]^{2}}{0.15 - x}\\\\\rm X &= 1.22 \times 10^{-4}\end{aligned}$

Percentage ionization can be shown as:

$\begin{aligned} \rm \% ionization &= \rm (\dfrac{[H+]}{[C_{6}H_{5}COOH]} ) \times 100\\\\\rm \% ionization &= (\dfrac{1.22\times 10^{-7}}{ 0.15}) \times 100\\\\\rm \% ionization &= 0.08\%\end{aligned}$

The percentage ionization for  sodium benzoate can be calculated as:

$\begin{aligned} 6.5 \times 10^{-5} &= \rm \dfrac{[X]^{2}}{(0.10 - X)}\\\\\rm X &= 2.5\times 10^{-3}\end{aligned}$

The ionization percentage:

$\begin{aligned}\rm \% ionization &= (\dfrac{2.5\times 10^{-3} }{0.10})\times 100\\\\&= 2.5\%\end{aligned}$

Therefore, the % ionization is different as benzoic acid is a weak acid while sodium benzoate is a salt.

Learn more about per cent ionization here:

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