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Buffer solution دکتر امید رجبی دانشیار گروه شیمی دارویی. شیمی عمومی. Buffer solutions are solutions that resist change in Hydronium ion and the hydroxide ion concentration (and consequently pH ) upon addition of small amounts of acid or base , or upon dilution. .

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Buffer solution l.jpg

Buffer solution دکتر امید رجبیدانشیار گروه شیمی دارویی

شیمی عمومی

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  • Buffer solutions are solutions that resist change in Hydronium ion and the hydroxide ion concentration (and consequently pH) upon addition of small amounts of acid or base, or upon dilution.

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Buffers drastically.

  • A buffer solution resists a change in pH when an acid or base is added.

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Components of a Buffer drastically.

A buffer solution

  • Contains a combination of acid-base conjugate pairs.

  • Contains a weak acid and a salt of the conjugate base of that acid.

  • Typically has equal concentrations of a weak acid and its salt.

  • May also contain a weak base and a salt with the conjugate acid.

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Buffer Action drastically.

  • The acetic acid/acetate buffer contains acetic acid (CH3COOH) and sodium acetate (CH3COONa).

  • The salt produces sodium and acetate ions.

    CH3COONa CH3COO- + Na+

  • The salt provides a higher concentration of the conjugate base CH3COO- than the weak acid.

    CH3COOH + H2O CH3COO- + H3O+

    small amount Large amount

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Function of the Weak Acid drastically.

  • The function of the weak acid is to neutralize a base. The acetate ion in the product adds to the available acetate.

    CH3COOH + OH- CH3COO- + H2O

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Function of the Conjugate Base drastically.

  • The function of the acetate ion CH3COO- (conjugate base) is to neutralize H3O+ from acids. The weak acid product adds to the weak acid available.

    CH3COO- + H3O+ CH3COOH + H2O

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Summary of Buffer Action drastically.

  • The weak acid in a buffer neutralizes base.

  • The conjugate base in the buffer neutralizes acid.

  • The pH of the solution is maintained.

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pH of a Buffer drastically.

  • The [H3O+] in the Ka expression is used to determine the pH of a buffer.

    Weak acid + H2O H3O+ + Conjugate base

    Ka = [H3O+][conjugate base]

    [weak acid]

    [H3O+] = Ka x [weak acid]

    [conjugate base]

    pH = -log [H3O+]

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Calculation of Buffer pH drastically.

The weak acid H2PO4- in a blood buffer H2PO4-/HPO42- has Ka = 6.2 x 10-8. What is the pH of the buffer if it is 0.20 M in both H2PO4- and HPO42-?

[H3O+] = Ka x [H2PO4-]


[H3O+] = 6.2 x 10-8 x [0.20 M] = 6.2 x 10-8

[0.20 M]

pH = -log [6.2 x 10-8] = 7.21

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The Henderson-Hasselbach Equation: A practical application drastically.

Let's begin our discussion of the Henderson-Hasselbach Equation with a continuation of the dissociation constant derivation.

should be a familiar relationship . If we rearrange our equation to provide the [H3O+] on the left side, we get

[H3O+] =Ka (HA)/(A-)

(Notice that the ionized species of the acid is now in the denominator.)

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Taking the negative log of both sides gives us easier numeric values to work with:

which can be rewritten

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more familiarly numeric values to work with:

Henderson-Hasselbach Equation