{"id":6085,"date":"2020-10-13T14:08:58","date_gmt":"2020-10-13T08:38:58","guid":{"rendered":"http:\/\/astan.lk\/al_virtualclassroom\/?p=6085"},"modified":"2020-10-13T14:08:21","modified_gmt":"2020-10-13T08:38:21","slug":"buffer-solutions","status":"publish","type":"post","link":"https:\/\/astan.lk\/al_virtualclassroom\/buffer-solutions\/","title":{"rendered":"Buffer solutions"},"content":{"rendered":"

\u2022 It can be shown either by calculation or using pH papers that pH of chemically pure water can be changed by 3 units by adding 1 cm3<\/sup> of 0.1 mol dm-3<\/sup> HCl or NaOH to 1.0 dm3<\/sup> of water. This shows that a small amount of an acid or an alkali can cause a drastic change in pH. But there are solutions\/systems that can resist such changes.<\/p>\n

\u2022 A buffer solution is a one which resists the change in pH on addition of a small amount of H+<\/sup> or OH–<\/sup> or water.<\/p>\n

\u2022 The following table shows how pH changes when the given volumes of a 0.1 mol dm-3<\/sup> NaOH solution are added to 50.00 cm3<\/sup> of a 0.1 mol dm-3<\/sup> CH3<\/sub>COOH solution.<\/p>\n

\"\"<\/a><\/p>\n

 <\/p>\n

\u2022 According to the above table, the change in pH in between the addition of 10.00 cm3<\/sup> and 25.00 cm3<\/sup>of NaOH solution is very small showing that in this range, the system is resistant to the change in pH brought about by the addition of NaOH. This action is called buffer action.<\/p>\n

 <\/p>\n

Buffer systems<\/h4>\n

\u2022 Solutions containing a weak acid and its conjugate base act as buffers.
\neg. CH3<\/sub>COOH and CH3<\/sub>COONa.<\/p>\n

\u2022 Solutions containing a weak base and its conjugate acid also act as buffers.
\neg. NH4<\/sub>OH and NH4<\/sub>Cl.<\/p>\n

 <\/p>\n

The buffer action of the CH3<\/sub>COOH and CH3<\/sub>COONa system.<\/h4>\n

CH3<\/sub>COONa(aq) \u2192 CH3-(aq) + Na+<\/sup>(aq) (Complete dissociation)
\nCH3<\/sub>COOH(aq) + H2<\/sub>O(l) \u21cc<\/span><\/span><\/span><\/span>\u00a0<\/span><\/b>CH3COO–<\/sup>(aq) + H3<\/sub>O+<\/sup>(aq) (Incomplete dissociation)<\/sub><\/p>\n

When H3<\/sub>O+<\/sup> ions are added to the system, they are removed by the CH3<\/sub>COO–<\/sup> ion forming weakly dissociated CH3<\/sub>COOH. Therefore, pH almost remains constant<\/p>\n

When a small amount of OH–<\/sup> ions are added to the system, they are immediately removed forming almost unionised water.<\/p>\n

OH–<\/sup>(aq) + H3<\/sub>O+<\/sup>(aq) \u21cc<\/span><\/span><\/span><\/span><\/span><\/b>\u00a02H2O(l)<\/p>\n

More CH3<\/sub>COOH ionises to replenish the lost H3<\/sub>O+<\/sup>, so pH remains almost constant.<\/p>\n

<\/h4>\n

The buffer action of the NH4<\/sub>OH(aq) and NH4<\/sub>Cl(aq) system.<\/h4>\n

NH4<\/sub>OH(aq) \u00a0 \u21cc<\/span><\/span><\/span><\/span><\/span><\/b> \u00a0NH4<\/sub>+<\/sup>(aq) + OH–<\/sup>(aq)
\nNH4<\/sub>Cl(aq) \u00a0\u2192 \u00a0 NH4<\/sub>+<\/sup>(aq) \u00a0+ Cl–<\/sup>(aq)<\/p>\n

When a small amount of an acid is added, H+<\/sup> ions are removed by the OH–<\/sup> forming water and more NH4<\/sub>OH ionises restoring the OH–<\/sup> ions. Hence, pH does not change widely. When an alkali is added, OH–<\/sup> ions combine with NH4<\/sub> +<\/sup> to form NH4<\/sub>OH bringing down the OH–<\/sup> concentration. Hence, pH approximately remains constant.<\/p>\n

 <\/p>\n

Henderson equation<\/h4>\n

Here is the dissociation equation for HA:<\/p>\n

HA \u21cc H+<\/sup> + A–<\/sup><\/p><\/blockquote>\n

From which, we write the Ka<\/sub> expression:<\/p>\n

<\/p><\/blockquote>\n

<\/p><\/blockquote>\n

take the negative log of each of the three terms in the above equation. They become:<\/p>\n

1) -log [H+<\/sup>]
\n2) – log Ka<\/sub>
\n3) -log ([HA] \/ [A\u00af])<\/p><\/blockquote>\n

However,<\/p>\n

1) this is the pH
\n2) this is the pKa<\/sub>
\n3) to get rid of the negative sign \u00a0flip the log term to get this: + log ([A\u00af] \/ [HA])<\/p><\/blockquote>\n

Inserting these last three items (the pH, the pKa<\/sub> and the rearranged log term), we arrive at the Henderson-Hasselbalch Equation:<\/p>\n

<\/p><\/blockquote>\n

Here is a common way the HH equation is presented in a textbook explanation:<\/p>\n

<\/p><\/blockquote>\n

Remember that, in a buffer, the two substances differ by only a proton. The substance with the proton is the acid and the substance without the proton is the salt.<\/p>\n

However, remember that the salt of a weak acid is a base (and the salt of a weak base is an acid).<\/p>\n

Consequently, another common way to write the Henderson equation is to substitute “base” for “salt form” (sometimes you will see “conjugate base” or “base form”). This is probably the most useful way to decribe the interactions between the acidic form (the HA) and the basic form (the A\u00af).<\/p>\n

Here it is:<\/p>\n

<\/p><\/blockquote>\n

 <\/p>\n

 <\/p>\n

The alternate form starts from the ionization equation for a generic base called B:<\/p>\n

B + H2<\/sub>O \u21cc HB+<\/sup> + OH\u00af<\/p><\/blockquote>\n

By the way, here is an example of the above generic equation, using ammonia:<\/p>\n

NH3<\/sub> + H2<\/sub>O \u21cc NH4<\/sub>+<\/sup> + OH\u00af<\/p><\/blockquote>\n

The B simply represents the entire base and HB+<\/sup> represents the substances with an additional H+<\/sup>.<\/p>\n

Next, we write the Kb<\/sub> expression for this reaction:<\/p>\n

<\/p><\/blockquote>\n

Next, we isolate the [OH\u00af] on the left-hand side of the equation:<\/p>\n

<\/p><\/blockquote>\n

We negative log each of the three terms of the above equation to give pOH, pKb and we flip the third term so as to make it an addition, not a subtraction.<\/p>\n

Here is the alternate form of the Henderson-Hasselbalch Equation, expressed in terms of pOH and pKb<\/sub>:<\/p>\n

<\/p><\/blockquote>\n

\u00a0<\/span><\/p><\/blockquote>\n","protected":false},"excerpt":{"rendered":"

\u2022 It can be shown either by calculation or using pH papers that pH of chemically pure water can be changed by 3 units by adding 1 cm3 of 0.1 mol dm-3 HCl or NaOH to 1.0 dm3 of water. 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