We would like to invite you to participate in the Analytical Challenge, a series of puzzles to entertain and challenge our readers. This special feature of “Analytical and Bioanalytical Chemistry” has established itself as a truly unique quiz series, with a new scientific puzzle published every other month. Readers can access the complete collection of published problems with their solutions on the ABC homepage at http://www.springer.com/abc . Test your knowledge and tease your wits in diverse areas of analytical and bioanalytical chemistry by viewing this collection.

In the present challenge, ‘dilution’ is the topic. And please note that there is a prize to be won (a Springer book of your choice up to a value of €100). Please read on…

Meet the challenge

Ostwald’s dilution law has become a staple of physical and general chemistry. It refers to the formula introduced in 1888 [1] by the Latvian-born chemist Friedrich Wilhelm Ostwald (1853–1932), one of the founders of physical chemistry and the 1909 Nobel Laureate for Chemistry [2]. Ostwald’s dilution law refers, in particular, to the relation between the dissociation constant, K a, and the degree of ionization, α, of a weak monoprotic acid, HL [3]. On the basis of reaction

$$ \mathrm{HL}={\mathrm{H}}^{+}+{\mathrm{L}}^{\hbox{--} } $$
(1)

we formulate the expression for dissociation constant K a of HL

$$ {K}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{L}}^{-}\right]}{\left[\mathrm{HL}\right]} $$
(2)

In this solution, we can establish the equations for concentration balance

$$ \left[\mathrm{HL}\right]+\left[{\mathrm{L}}^{\hbox{--}}\right]=C $$
(3)

and charge balance [4, 5]

$$ \left[{\mathrm{H}}^{+}\right]-\left[{\mathrm{OH}}^{\hbox{--}}\right]-\left[{\mathrm{L}}^{\hbox{--}}\right]=0 $$
(4)

The degree of dissociation (ionization) of HL is defined by formula

$$ \alpha =\frac{\left[{\mathrm{L}}^{-}\right]}{C} $$
(5)

Applying Eqs. (3) and (4) in (2), we obtain

$$ {K}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left(\left[{\mathrm{H}}^{+}\right]-\left[{\mathrm{OH}}^{-}\right]\right)}{C-\left[{\mathrm{H}}^{+}\right]+\left[{\mathrm{OH}}^{-}\right]} $$
(6)

On the basis of (4) and (5), at [H+] >> [OH−], one can write

$$ \left[{\mathrm{H}}^{+}\right]=\left[{\mathrm{L}}^{\hbox{--}}\right]=\upalpha \cdot C $$
(7)

and from (2) or (6) we have the well-known formula

$$ {K}_{\mathrm{a}}=\frac{C\cdot {\alpha}^2}{1-\alpha } $$
(8)

expressing the Ostwald’s dilution law. The use of the word dilution suggests that the law is particularly valid for diluted solutions. This challenge seeks to find if this is indeed the case.

The challenge

The limiting value for α at extremely diluted solutions can be obtained by transforming Eq. (8) into quadratic equation

$$ C\cdot {\alpha}^2+{K}_{\mathrm{a}}\cdot \alpha \hbox{--} {K}_{\mathrm{a}}=0 $$
(9)

From here we obtain

$$ \alpha =\frac{K_{\mathrm{a}}}{2C}\left(\sqrt{1+4\cdot C/{K}_{\mathrm{a}}}-1\right) $$
(10)

Applying the approximation \( \sqrt{1+4C/{K}_{\mathrm{a}}}\approx 1+2C/{K}_{\mathrm{a}} \), valid for extremely diluted solutions with 4C << K a, we have

$$ \alpha \left(C\to 0\right)=\underset{C\to 0}{ \lim}\frac{K_{\mathrm{a}}}{2C}\left(\sqrt{1+4\cdot C/{K}_{\mathrm{a}}}-1\right)\approx \frac{K_{\mathrm{a}}}{2C}\cdot \frac{2C}{K_{\mathrm{a}}}=1 $$
(11)

The above equation suggests that all weak acids dissociate completely at infinite dilution, irrespective of the K a value. This statement, however, is invalid. Consider, for example, cyanic acid, HCN with pK a = 9.2. In large dilutions, the pH of HCN solutions will be near that of pure water, and HCN will be almost completely undissociated.

Can you find the proper expression for calculating the degree of dissociation?

Calculate the degree of dissociation of acetic acid under infinite dilution.