#ionic equilibrium

A student dissolved a certain amount of a weak acid (HA) in some water. She then titrated this solution against NaOH of some unknown concentration. When she had added 5mL of NaOH the pH of the solution was found to be 5. It took a total of 20mL of NaOH to completely neutralize the acid solution she had prepared. What is the acid dissociation constant (\(K_a\)) of the weak acid, HA?

Answer

let us assume the total moles of the acid in solution as “x”.

20mL of NaOH is required to completely neutralize the acid solution. This means that 5mL of NaOH will be able to neutralize \(\frac{1}{4}\times x\) moles of the acid taken and produce \(\frac{1}{4} \times x\) moles of the salt of this acid with a strong base. The resulting solution will be an acidic buffer.

pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

\[pH = pK_a + log\dfrac{[A^-]}{[HA]}\]

We have \(\frac{3}{4}\times x\) moles of the acid (\([HA]\)) remaining and \(\frac{1}{4} \times x\) moles of the salt (\(A^-\)). The pH of the solution was 5 at this point of the titration.

So,

\[pH = pK_a + log\dfrac{[A^-]}{[HA]}\]

\[5 = pK_a + log\dfrac{\frac{1}{4}}{\frac{3}{4}}\]

\[5 = pK_a + log\dfrac{1}{3}\]

\[5 = pK_a - log3\]

\[5 = pK_a - 0.477\]

\[pKa = 5 + 0.477\]

\[pKa = 5.477\]

\[-log K_a = 5.477\]

At a given temperature pH of all solutions will be falling within a given range. This range depends on the extent to which water is ionized at that given temperature.

Water, being a weak electrolyte, will be partially ionized. The partial ionization of water can be represented as below;

\[H_2O \rightleftharpoons H^+ + HO^-\]

At \(25^o C\), the degree of dissociation (\(\alpha\)) of water is \(1.8 \times 10^{-9}\). This means that for every mole of water taken \(1.8 \times 10^{-9}\) moles of water is ionized.

The acid dissociation constant of water (\(K_a\)) can be written as,

\(K_a = \frac {[H^+][HO^-]}{[H_2O]} = \frac{C \alpha \times C \alpha}{C(1-\alpha)}\)

where C is the concentration of water and \(\alpha\) is the degree of dissociation of water.

The concentration of water is \(\frac {1000 g/L}{18 g/mol} \approx 55.5 mol/L \)

As \(\alpha \ll 1\), we can neglect it with respect to 1. Neglecting it with respect to 1 and substituting the values of C and \(\alpha\) in the above equation for \(K_a\) we get,

\(K_a = C\alpha^2 = \frac{1000}{18} \times 1.8 \times 1.8 \times 10^{-18} = 1.8 \times 10^{-16}\)

We can now define another constant \(K_w\), the ionic product of water as,

\(K_w = K_a \times [H_2O] = [H^+][HO^-]\)

\(K_w = 1.8 \times 10^{-16} \times \frac{1000}{18} = 10^{-14}\)

\(K_w = 10^{-14} = [H^+][HO^-] = [H^+]^2 (because [H^+] = [HO^-])\)

So, \([H^+]^2 = 10^{-7}\)

\(pH = -log[H^+] = -log 10^{-7} = 7\)

pH of water at \(25^oC\) is 7. This will be the neutral pH as the \([H^+] = [HO^-]\). If the \([H^+] > [HO^-]\), then the value of pH will become less than 7 and if the \([H^+] < [HO^-]\), then the value of pH will become more than 7. To determine the lower and upper limit of the pH value at this temperature and thus the pH scale at this temperature we have to down by 7 units and up by 7 units because 7 will be the mid point of the scale. Thus, the pH scale at \(25^oC\) will be 0 to 14.

Let us say that, at \(125^oC\), \(K_w\) of water is \(1.0 \times 10^{-12}\). What would be the pH of pure water at this temperature and if a solution with pH = 7, is available at this temperature, will it be acidic, basic or neutral?

\(K_w = 1.0 \times 10^{-12} = [H^+][HO^-] = [H^+]^2 (because [H^+] = [HO^-])\)

So, \([H^+]^2 = 10^{-6}\)

\(pH = -log[H^+] = -log 10^{-6} = 6\)

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of \(5.7^oC\) was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (\(- 57 kJ mol^{-1}\)), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2), 100 mL of 2.0 M acetic acid (\(K_a = 2.0 \times 10^{-5}\)) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of \(5.6^oC\) was measured. (Consider heat capacity of all solutions as \(4.2 J g^{-1} K^{-1}\) and density of all solutions as \(1.0 g mL^{-1}\))

Enthalpy of dissociation (\(in \ kJ mol^{-1}\)) of acetic acid obtained from Expt. 2 is A) 1.0 B) 10.0 C) 24.5 D) 51.4

Selected References to Signaling and Metabolic Pathways in Leaders in Pharmaceutical Intelligence

Larry H. Bernstein, MD, FCAP, Reporter and Curator

Leaders in Pharmaceutical Intelligence

http://pharmaceuticalintelligence.com/8-10-2014/Selected References to Signaling and Metabolic Pathways in Leaders in Pharmaceutical Intelligence

  This is an added selection of articles in Leaders in Pharmaceutical Intelligence after the third portion of the discussion in a series of articles that began…

Equal volumes of two solutions of HCl are mixed. One solution has pH = 1 while the other has pH = 5. What will the pH of the solution obtained be?

Solution:

Let the volume of the solutions mixed = “V” mL

So, the volume of the final solution obtained after mixing = “2V” mL

For solution with pH = 1,

Concentration of H+ = 10-1 M

Moles of H+ = 10-1 V

For solution with pH = 5,

Concentration of H+ = 10-5M

Moles of H+ = 10-5 V

For the Final Solution,

Total moles of H+ in the final solution = (10-1 V) + (10-5 V) ≈ 10-1 V

Concentration of H+ in the final solution = (10-1 V) / 2V = 0.05 M