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Preparation of buffers

Preparation of buffers. Definition. A 1M solution of sulfuric acid contains 98.06 g of sulfuric acid in 1 liter of total solution. "mole" is an expression of amount "molarity" is an expression of concentration. "Millimolar", mM, millimole/L. A millimole is 1/1000 of a mole .

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Preparation of buffers

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  1. Preparation of buffers

  2. Definition • A 1M solution of sulfuric acid contains 98.06 g of sulfuric acid in 1 liter of total solution. • "mole" is an expression of amount • "molarity" is an expression of concentration. • "Millimolar", mM, millimole/L. • A millimole is 1/1000 of a mole. • "Micromolar", µM, µmole/L. • A µmole is 1/1,000,000 of a mole.

  3. FORMULA HOW MUCH SOLUTE IS NEEDED FOR A SOLUTION OF A PARTICULAR MOLARITY AND VOLUME? • (g solute ) X (mole) X (L) = g solute needed • 1 mole L • or • FW X molarity x volume = g solute needed

  4. EXAMPLE How much solute is required to make 300 mL of 0.8 M CaCl2? • (111.0 g) (0.8 mole) (0.3 L) = 26.64 g • mole L

  5. Buffers • Buffers are the solutions which resist changes in pH when small amounts of acid or alkali is added to them. • A buffer is a pair of weak acid and its salt. • Buffers are of main importance in regulating the pH of the body fluids and tissues • Many biochemical reactions including those catalyzed by enzymes require pH control which is provided by buffers

  6. Buffers • Mammalian tissues in the resting state have a pH of about 7.4 • In order to maintain the required pH in an in vitro biochemical experiment a buffer is always used • The pH of a buffer is given by Handerson-Hasselbalch equation • pH= pKa + log [A-] • [HA] • pKa= -log Ka • Ka is the dissociation constant of the acid • [A-] is the concentration of the base • [HA] is the concentration of the acid

  7. Buffers • A buffer consisting of a mixture of the weak acid (CH3COOH) and its salt (CH3COONa+) Will undergo the following changes on the addition of acid or base: If the base is added (OH) to the solution it will be buffered by the following reaction with acetic acid: CH3COOH + OHCH3COO- + H2O So the pH will not change significantly

  8. Buffers • If acid (H+) is added, it will be buffered by another reaction, this time using the salt (CH3COO-) CH3COO- + H+  CH3COOH • The pH will not alter significantly because the CH3COOH formed is a weak acid Addition of more base increases A- and decreases (HA) and this doesn’t alter the pH much until [A-]>>>>[HA] • The buffering power is greatest when pH=pKa i.e. when the acid and the salt are at the same concentration

  9. Method • You are provided with 0.1M sodium acetate and 0.1M acetic acid solution • Prepare 50ml of the buffer solution by mixing the two standard solutions • Use the following information to calculate the composition of the buffer: pH of the buffer 4.86 pKa of acetic acid at 25C 4.76

  10. Four practical methods to prepare a buffer: • Prepare a buffer composed of an acid and its salt by adding a strong base (e.g. NaOH) to a weak acid (e.g. Acetic acid) until the required pH is obtained. •  If the other form of buffer is available (in this case sodium acetate), a strong acid is added (e.g. HCl) until the required pH is obtained. • CH3COONa+HClCH3COOH+NaCl • So acetate buffer is formed(CH3COOH/CH3COONa)

  11. The Third method: • Using the buffer pKa , calculate the amounts (in moles) of acid/salt or base/salt present in the buffer at the desired pH. • If both forms (i.e., the acid and the salt) are available, convert the amount required from moles to grams ,using the molecular weight of that component, and then weigh out the correct amounts of  both forms. Or convert moles to volume if the stock is available in the liquid form. • The fourth method: • Find a table of the correct amounts of acid/salt or base/salt required for different pH's • Dissolve the components in slightly less water than is required for the final solution volume.  • Check that the pH and correct if necessary. • Add water to the final volume.

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