Protolytic Equilibria in Pharmaceutical Solutions

Step-by-Step Problem Solving for First-Year Pharmacy Students
Pharmaceutical Chemistry Department
1

Hydrogen Ion Concentration in Blood

The normal pH of human blood is about 7.4. Calculate the hydrogen ion concentration [H⁺] corresponding to this pH value.

Step-by-Step Solution:

Recall the pH formula:
\(\text{pH} = -\log_{10}[\text{H}^+]\)
Rearrange the formula to solve for [H⁺]:
\([\text{H}^+] = 10^{-\text{pH}}\)
Substitute the given pH value:
\([\text{H}^+] = 10^{-7.4}\)
Calculate the hydrogen ion concentration:
\([\text{H}^+] = 3.98 \times 10^{-8} \text{mol/L}\)
The narrow pH range of blood (7.35-7.45) is critical for maintaining proper physiological function. Even small deviations can lead to serious health issues like acidosis or alkalosis.
[H⁺] = 3.98 × 10⁻⁸ mol/L
2

Urine pH Diagnosis

The pH of human urine normally varies in the range of 4.8–8.0. The patient urine analysis [H⁺] = 5.8×10⁻⁴. Diagnose the patient – acidosis, alkalosis or normal?

Step-by-Step Solution:

Calculate pH from [H⁺]:
\(\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(5.8 \times 10^{-4})\)
Break down the logarithm calculation:
\(\text{pH} = -(\log_{10}5.8 + \log_{10}10^{-4})\)
\(\text{pH} = -(0.763 - 4) = 3.237\)
Compare with normal range:

Normal urine pH range: 4.8–8.0

Calculated pH: 3.24

Diagnose the patient:

Since 3.24 < 4.8, the urine is abnormally acidic.

Acidic urine (pH < 4.8) may indicate metabolic acidosis, uncontrolled diabetes, or certain kidney disorders. This is important for clinical diagnosis and treatment planning.
The patient has acidosis (pH = 3.24, below normal range)
3

Breast Milk pH Diagnosis

The pOH of a woman's breast milk normally varies in the range of 6.5–7.1. The female patient's breast milk analysis [H⁺] = 4.3×10⁻⁵. Diagnose the patient – acidosis, alkalosis or normal?

Step-by-Step Solution:

Calculate pH from [H⁺]:
\(\text{pH} = -\log_{10}(4.3 \times 10^{-5}) = 4.37\)
Calculate pOH using the relationship:
\(\text{pH} + \text{pOH} = 14\)
\(\text{pOH} = 14 - \text{pH} = 14 - 4.37 = 9.63\)
Compare with normal range:

Normal breast milk pOH range: 6.5–7.1

Calculated pOH: 9.63

Diagnose the patient:

Since 9.63 > 7.1, the breast milk is abnormally acidic (low pH, high pOH).

Abnormal breast milk pH can affect its nutritional value and antimicrobial properties. This calculation demonstrates how pH/pOH relationships are used in clinical diagnostics for various body fluids.
The patient has acidosis (pOH = 9.63, above normal range)
4

Bile pH Diagnosis

The pH of human bile normally varies in the range of 8.0–8.5. The patient's bile [OH⁻] analysis is 0.00012 mol/L. Diagnose the patient – acidosis, alkalosis or normal?

Step-by-Step Solution:

Calculate pOH from [OH⁻]:
\(\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(0.00012)\)
\(\text{pOH} = -\log_{10}(1.2 \times 10^{-4}) = 3.92\)
Calculate pH using the relationship:
\(\text{pH} = 14 - \text{pOH} = 14 - 3.92 = 10.08\)
Compare with normal range:

Normal bile pH range: 8.0–8.5

Calculated pH: 10.08

Diagnose the patient:

Since 10.08 > 8.5, the bile is abnormally alkaline.

Bile pH is crucial for fat digestion and absorption. Abnormally high pH may indicate liver or gallbladder issues, which can affect drug metabolism and nutrient absorption.
The patient has alkalosis (pH = 10.08, above normal range)
5

Synovial Fluid pH Diagnosis

The pOH of the synovial fluid of human knee joints normally varies in the range of 6.4–6.7. The patient's synovial fluid [H⁺] analysis is 3.98×10⁻⁸ mol/L. Diagnose the patient – acidosis, alkalosis or normal?

Step-by-Step Solution:

Calculate pH from [H⁺]:
\(\text{pH} = -\log_{10}(3.98 \times 10^{-8}) = 7.40\)
Calculate pOH using the relationship:
\(\text{pOH} = 14 - \text{pH} = 14 - 7.40 = 6.60\)
Compare with normal range:

Normal synovial fluid pOH range: 6.4–6.7

Calculated pOH: 6.60

Diagnose the patient:

Since 6.4 ≤ 6.60 ≤ 6.7, the synovial fluid pH is within normal range.

Synovial fluid pH is an important indicator of joint health. Abnormal pH levels can indicate inflammation or infection in the joint, which is relevant for diagnosing conditions like arthritis.
The patient is normal (pOH = 6.60, within normal range)
6

Small Intestine pH Change

The normal pH of the human small intestine fluid is about 7.2. If the absorption of fatty acids is impaired, the pH decreases by 2 units compared to normal. Calculate how many times the concentration of protons [H⁺] will change.

Step-by-Step Solution:

Calculate normal [H⁺]:
\([\text{H}^+]_{\text{normal}} = 10^{-\text{pH}} = 10^{-7.2} = 6.31 \times 10^{-8} \text{mol/L}\)
Calculate new pH:

pH decreases by 2 units: 7.2 - 2 = 5.2

Calculate new [H⁺]:
\([\text{H}^+]_{\text{new}} = 10^{-5.2} = 6.31 \times 10^{-6} \text{mol/L}\)
Calculate the change factor:
\(\text{Change factor} = \frac{[\text{H}^+]_{\text{new}}}{[\text{H}^+]_{\text{normal}}} = \frac{6.31 \times 10^{-6}}{6.31 \times 10^{-8}} = 100\)
Alternative calculation using pH difference:
\(\text{Change factor} = 10^{\Delta \text{pH}} = 10^{2} = 100\)
This demonstrates the logarithmic nature of the pH scale - a 2-unit decrease in pH represents a 100-fold increase in [H⁺]. This is critical for understanding how small pH changes can significantly impact drug absorption and metabolism in the gastrointestinal tract.
The proton concentration increases by 100 times
7

Dihydrogen Phosphate Buffer System

The normal pH of the dihydrogen phosphate buffer system of human blood is 7.41. The concentration of sodium dihydrogen phosphate is 0.15 mol/L. Calculate the Ka and pKa of acid of the dihydrogen phosphate buffer system.

Step-by-Step Solution:

Identify the buffer components:

Acid: H₂PO₄⁻ (dihydrogen phosphate)

Conjugate base: HPO₄²⁻ (hydrogen phosphate)

Given: pH = 7.41, [H₂PO₄⁻] = 0.15 mol/L

Use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{base}]}{[\text{acid}]}\right)\)
Assume equal concentrations of acid and base:

In a buffer system at its optimal pH, [base] = [acid]

\(\log_{10}\left(\frac{[\text{base}]}{[\text{acid}]}\right) = \log_{10}(1) = 0\)
Calculate pKa:
\(\text{pH} = \text{p}K_a\)
\(\text{p}K_a = 7.41\)
Calculate Ka:
\(K_a = 10^{-\text{p}K_a} = 10^{-7.41} = 3.89 \times 10^{-8}\)
The dihydrogen phosphate buffer system is crucial for maintaining blood pH. Understanding its pKa helps in designing pharmaceutical formulations that won't disrupt this important physiological buffer.
Ka = 3.89 × 10⁻⁸, pKa = 7.41
8

Hemoglobin Buffer System

The normal pH of the hemoglobin buffer system of human blood is 7.39. The concentration of the protonated form of hemoglobin HHb is 0.06 mol/L. Calculate the Ka and pKa of acid of the hemoglobin buffer system.

Step-by-Step Solution:

Identify the buffer components:

Acid: HHb (protonated hemoglobin)

Conjugate base: Hb⁻ (deprotonated hemoglobin)

Given: pH = 7.39, [HHb] = 0.06 mol/L

Use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Hb}^-]}{[\text{HHb}]}\right)\)
Assume equal concentrations of acid and base:

At the optimal buffer pH, [Hb⁻] = [HHb]

\(\log_{10}\left(\frac{[\text{Hb}^-]}{[\text{HHb}]}\right) = \log_{10}(1) = 0\)
Calculate pKa:
\(\text{pH} = \text{p}K_a\)
\(\text{p}K_a = 7.39\)
Calculate Ka:
\(K_a = 10^{-\text{p}K_a} = 10^{-7.39} = 4.07 \times 10^{-8}\)
The hemoglobin buffer system is essential for maintaining blood pH and facilitating oxygen transport. Understanding its pKa is crucial for developing drugs that won't disrupt this vital physiological process.
Ka = 4.07 × 10⁻⁸, pKa = 7.39
9

Dihydrogen Phosphate Ion pKa

The dissociation degree of the weak electrolyte dihydrogen phosphate ion contained in human blood is 0.15. Calculate the pKa of the dihydrogen phosphate ion if the concentration of NaH₂PO₄ is 0.2 mol/L.

Step-by-Step Solution:

Recall the relationship between dissociation degree (α) and Ka:
\(K_a = \frac{c\alpha^2}{1-\alpha}\)

Where c = initial concentration, α = dissociation degree

Substitute the given values:
\(K_a = \frac{0.2 \times (0.15)^2}{1-0.15}\)
Calculate Ka:
\(K_a = \frac{0.2 \times 0.0225}{0.85} = \frac{0.0045}{0.85} = 5.29 \times 10^{-3}\)
Calculate pKa:
\(\text{p}K_a = -\log_{10}(K_a) = -\log_{10}(5.29 \times 10^{-3})\)
\(\text{p}K_a = 2.28\)
This calculation demonstrates how to determine acid strength from experimental data. The pKa value is crucial for understanding drug behavior in physiological environments and designing effective pharmaceutical formulations.
pKa = 2.28