Buffer Solutions in Pharmaceutical Chemistry

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

Phosphate Buffer for Liver Enzyme Studies

When studying the activity of the liver enzyme alanine aminotransferase ALT, a phosphate buffer solution is used. To prepare such a solution, 840 ml of 0.1 M sodium hydrogen phosphate is mixed with 160 ml of 0.1 M sodium dihydrogen phosphate solution. Calculate the pH of this solution (pKa(H₂PO₄⁻) = 7.2).

Step-by-Step Solution:

Calculate moles of each component:
\(\text{Moles of } \text{HPO}_4^{2-} = 0.840 \text{L} \times 0.1 \text{M} = 0.084 \text{mol}\)
\(\text{Moles of } \text{H}_2\text{PO}_4^- = 0.160 \text{L} \times 0.1 \text{M} = 0.016 \text{mol}\)
Apply the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right)\)
Calculate the ratio of concentrations:

Since the total volume is the same for both components, we can use moles directly:

\(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = \frac{0.084}{0.016} = 5.25\)
Calculate the pH:
\(\text{pH} = 7.2 + \log_{10}(5.25) = 7.2 + 0.72 = 7.92\)
This phosphate buffer is crucial for maintaining the optimal pH for liver enzyme activity studies. The calculated pH (7.92) is within the physiological range where ALT enzyme functions properly, ensuring accurate laboratory results for liver function tests.
The pH of the phosphate buffer solution is 7.92
2

Urine Buffer System Ratios

The pH of human urine normally varies in the range of 4.8–8.0. At what ratios of the volumes of sodium dihydrogen phosphate and sodium hydrogen phosphate is the urine pH 4.5 and 8.5? (pKa(H₂PO₄⁻) = 7.2)

Step-by-Step Solution:

For pH = 4.5 (acidic urine):
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right)\)
\(4.5 = 7.2 + \log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right)\)
Solve for the ratio:
\(\log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right) = 4.5 - 7.2 = -2.7\)
\(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = 10^{-2.7} = 0.002\)

Since concentrations are equal (0.1 M), volume ratio = 0.002:1

For pH = 8.5 (alkaline urine):
\(8.5 = 7.2 + \log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right)\)
\(\log_{10}\left(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}\right) = 1.3\)
\(\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = 10^{1.3} = 20\)

Volume ratio = 20:1

Interpret the results:

For pH 4.5: 1 part sodium hydrogen phosphate to 500 parts sodium dihydrogen phosphate

For pH 8.5: 20 parts sodium hydrogen phosphate to 1 part sodium dihydrogen phosphate

The phosphate buffer system is important in urine for maintaining pH within physiological limits. These calculations help understand how the body regulates urine pH, which is crucial for kidney function and drug excretion.
For pH 4.5: Volume ratio = 0.002:1
For pH 8.5: Volume ratio = 20:1
3

Blood Alkali Buffer Capacity

Calculate the alkali buffer capacity of blood, if, when adding 1 ml of 0.05 M KOH to 100 ml of blood, the blood pH increases from 7.36 to 7.6.

Step-by-Step Solution:

Understand the buffer capacity formula:
\(\text{Buffer capacity} = \frac{\text{moles of base added}}{\Delta \text{pH} \times \text{volume of solution (L)}}\)
Calculate moles of KOH added:
\(\text{Moles of KOH} = 0.001 \text{L} \times 0.05 \text{M} = 5 \times 10^{-5} \text{mol}\)
Calculate the pH change:
\(\Delta \text{pH} = 7.6 - 7.36 = 0.24\)
Calculate buffer capacity:
\(\text{Buffer capacity} = \frac{5 \times 10^{-5}}{0.24 \times 0.1} = 0.00208 \text{mol/L·pH}\)
\(= 2.08 \text{mmol/L·pH}\)
Blood buffer capacity is critical for maintaining physiological pH. This value (2.08 mmol/L·pH) indicates how well blood can resist pH changes when bases are added, which is essential for understanding acid-base balance in clinical practice.
The alkali buffer capacity of blood is 2.08 mmol/L·pH
4

Hemoglobin Forms in Alkalosis

A patient in a state of metabolic alkalosis has the blood pH = 7.6. Calculate for this patient the ratio of the concentrations of deprotonated and protonated forms of hemoglobin in the blood, if pKa(HHb) = 8.2.

Step-by-Step Solution:

Use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Hb}^-]}{[\text{HHb}]}\right)\)
Substitute the known values:
\(7.6 = 8.2 + \log_{10}\left(\frac{[\text{Hb}^-]}{[\text{HHb}]}\right)\)
Solve for the ratio:
\(\log_{10}\left(\frac{[\text{Hb}^-]}{[\text{HHb}]}\right) = 7.6 - 8.2 = -0.6\)
\(\frac{[\text{Hb}^-]}{[\text{HHb}]} = 10^{-0.6} = 0.251\)
Interpret the result:

The ratio of deprotonated to protonated hemoglobin is 0.251:1

This means there's less deprotonated hemoglobin (Hb⁻) than protonated hemoglobin (HHb)

In alkalosis (high blood pH), hemoglobin tends to hold onto protons more tightly, which affects its oxygen-binding capacity. This ratio is important for understanding how acid-base imbalances impact oxygen transport in the blood.
The ratio of [Hb⁻]/[HHb] = 0.251
5

Blood pH After Adding HCl

The acid buffer capacity of the blood is 50 mmol/L. Calculate the blood pH value after addition of 4.6 ml of 0.05 M HCl to 100 ml of blood with pH = 7.36.

Step-by-Step Solution:

Calculate moles of HCl added:
\(\text{Moles of HCl} = 0.0046 \text{L} \times 0.05 \text{M} = 2.3 \times 10^{-4} \text{mol}\)
Convert to concentration in blood:
\([\text{HCl}] = \frac{2.3 \times 10^{-4} \text{mol}}{0.1 \text{L}} = 0.0023 \text{mol/L} = 2.3 \text{mmol/L}\)
Use buffer capacity formula:
\(\text{Buffer capacity} = \frac{\text{acid added}}{|\Delta \text{pH}|}\)
\(|\Delta \text{pH}| = \frac{\text{acid added}}{\text{buffer capacity}} = \frac{2.3}{50} = 0.046\)
Calculate the new pH:

Since acid is added, pH decreases:

\(\text{New pH} = 7.36 - 0.046 = 7.314\)
The acid buffer capacity (50 mmol/L) indicates blood's ability to resist pH changes when acids are added. This calculation shows how even small acid additions can affect blood pH, highlighting the importance of buffer systems in maintaining physiological pH balance.
The blood pH after adding HCl is 7.314
6

NaHCO₃ Volume for Target pH

Determine the volume of 0.5 mol/L NaHCO₃ solution that must be added to 50 ml of 0.3 mol/L H₂CO₃ solution so that pH = 4.65 is established in the system, if Ka(H₂CO₃) = 1.7×10⁻⁴.

Step-by-Step Solution:

Calculate pKa:
\(\text{p}K_a = -\log_{10}(1.7 \times 10^{-4}) = 3.77\)
Use the Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right)\)
\(4.65 = 3.77 + \log_{10}\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right)\)
Solve for the ratio:
\(\log_{10}\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) = 0.88\)
\(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = 10^{0.88} = 7.59\)
Set up the equation for moles:

Let V be the volume of NaHCO₃ added (in L)

\(\frac{0.5V}{0.05 \times 0.3 - 0.5V} = 7.59\)
\(0.5V = 7.59(0.015 - 0.5V)\)
Solve for V:
\(0.5V = 0.11385 - 3.795V\)
\(4.295V = 0.11385\)
\(V = 0.0265 \text{L} = 26.5 \text{mL}\)
This calculation is essential for preparing buffer solutions with specific pH values, which is critical in pharmaceutical formulations where precise pH control is needed for drug stability and efficacy.
26.5 mL of 0.5 mol/L NaHCO₃ solution must be added
7

Protein Buffer Capacity Analysis

For the protein (glycine) blood system determine whether the acid or alkali buffer capacity is greater, if it is known that when adding 5 ml of 0.1 M HCl to 10 ml of glycine buffer, its pH decreased by 1.3, and when adding 2 ml 0.25 M NaOH pH increased by 2.4.

Step-by-Step Solution:

Calculate acid buffer capacity:
\(\text{Acid buffer capacity} = \frac{\text{moles of acid added}}{|\Delta \text{pH}| \times \text{volume (L)}}\)
\(= \frac{0.005 \text{L} \times 0.1 \text{M}}{1.3 \times 0.01 \text{L}} = \frac{0.0005}{0.013} = 0.0385 \text{mol/L·pH}\)
Calculate alkali buffer capacity:
\(\text{Alkali buffer capacity} = \frac{\text{moles of base added}}{\Delta \text{pH} \times \text{volume (L)}}\)
\(= \frac{0.002 \text{L} \times 0.25 \text{M}}{2.4 \times 0.01 \text{L}} = \frac{0.0005}{0.024} = 0.0208 \text{mol/L·pH}\)
Compare the buffer capacities:

Acid buffer capacity = 0.0385 mol/L·pH

Alkali buffer capacity = 0.0208 mol/L·pH

Acid buffer capacity > Alkali buffer capacity

Interpret the result:

The protein (glycine) blood system has greater acid buffer capacity than alkali buffer capacity.

This asymmetry in buffer capacity is common in biological systems. The greater acid buffer capacity helps the body better handle acid loads, which is important for understanding how protein buffers protect against acidosis in clinical practice.
Acid buffer capacity is greater than alkali buffer capacity
8

Ammonium Buffer pH Calculation

Calculate the pH of an ammonium buffer of urine containing 15 ml of 0.5 M NH₄Cl and 35 ml of 0.85 M NH₄OH after adding 5 ml of 0.1 M NaOH (Ka(NH₄OH) = 5.62×10⁻¹⁰).

Step-by-Step Solution:

Calculate initial moles:
\(\text{Moles of } \text{NH}_4^+ = 0.015 \text{L} \times 0.5 \text{M} = 0.0075 \text{mol}\)
\(\text{Moles of } \text{NH}_3 = 0.035 \text{L} \times 0.85 \text{M} = 0.02975 \text{mol}\)
Calculate pKa and pKb:
\(\text{p}K_a = -\log_{10}(5.62 \times 10^{-10}) = 9.25\)
\(\text{p}K_b = 14 - 9.25 = 4.75\)
Calculate moles after adding NaOH:
\(\text{Moles of NaOH added} = 0.005 \text{L} \times 0.1 \text{M} = 0.0005 \text{mol}\)

NaOH reacts with NH₄⁺ to form NH₃:

\(\text{New moles of } \text{NH}_4^+ = 0.0075 - 0.0005 = 0.0070 \text{mol}\)
\(\text{New moles of } \text{NH}_3 = 0.02975 + 0.0005 = 0.03025 \text{mol}\)
Use Henderson-Hasselbalch equation:
\(\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{NH}_3]}{[\text{NH}_4^+]}\right)\)
\(\text{pH} = 9.25 + \log_{10}\left(\frac{0.03025}{0.0070}\right)\)
\(\text{pH} = 9.25 + \log_{10}(4.32) = 9.25 + 0.635 = 9.885\)
Ammonium buffers are important in urine for maintaining pH balance. This calculation shows how buffer systems respond to base addition, which is critical for understanding renal acid-base regulation and drug excretion mechanisms.
The pH of the ammonium buffer after adding NaOH is 9.89