# 14.6 Buffers  (Page 3/9)

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$\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)\phantom{\rule{0.2em}{0ex}}=9.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M$

The concentration of NaOH is:

$\frac{9.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M\phantom{\rule{0.2em}{0ex}}\text{NaOH}}{0.101\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}=9.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}M$

The pOH of this solution is:

$\text{pOH}=\text{−log}\left[{\text{OH}}^{\text{−}}\right]\phantom{\rule{0.2em}{0ex}}=\text{−log}\left(9.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\phantom{\rule{0.2em}{0ex}}=3.01$

The pH is:

$\text{pH}=14.00\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\text{pOH}=10.99$

The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).

Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 $×$ 10 −5 M HCl solution from 4.74 to 3.00.

Initial pH of 1.8 $×$ 10 −5 M HCl; pH = −log[H 3 O + ] = −log[1.8 $×$ 10 −5 ] = 4.74
Moles of H 3 O + in 100 mL 1.8 $×$ 10 −5 M HCl; 1.8 $×$ 10 −5 moles/L $×$ 0.100 L = 1.8 $×$ 10 −6
Moles of H 3 O + added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L $×$ 0.0010 L = 1.0 $×$ 10 −4 moles; final pH after addition of 1.0 mL of 0.10 M HCl:

$\text{pH}=\text{−log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\text{−log}\left(\phantom{\rule{0.2em}{0ex}}\frac{\text{total moles}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}}{\text{total volume}}\right)\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}=\text{−log}\left(\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{mol}+1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{mol}}{101\phantom{\rule{0.2em}{0ex}}\text{mL}\left(\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{L}}{1000\phantom{\rule{0.2em}{0ex}}\text{mL}}\right)\phantom{\rule{0.2em}{0ex}}}\right)\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}=3.00$

If we add an acid or a base to a buffer that is a mixture of a weak base and its salt, the calculations of the changes in pH are analogous to those for a buffer mixture of a weak acid and its salt.

## Buffer capacity

Buffer solutions do not have an unlimited capacity to keep the pH relatively constant ( [link] ). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion.

The buffer capacity    is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.

## Selection of suitable buffer mixtures

There are two useful rules of thumb for selecting buffer mixtures:

1. A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. [link] shows an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.
2. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.

The reaction of aceto nitrile with propane in the presence of the acid
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Manish
transforming reactants to product(s)
Andre
process
Andre
Example of Lewis acid
Example of Lewis acid
Chidera
Chlorine
Anything with an empty orbital... the hydrogen ion is the most common example. BH3 is the typical example, but any metal in a coordination complex can be considered a Lewis acid.
Eszter
okay thanks
Jovial
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Soni
aluminium and sulphur react to give aluminium sulphide how many grams of Al are required to produce 100g of aluminium sulphide?
Soni
2Al+3S=Al2S3
galina
m(Al)=100×27×2/150=36g
galina
150 comes from?
Soni
thank you very much
Soni
molar mass of Al2S3
galina
150.158
thiru
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matter simply converts to pure energy
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HF is
HF
Duku
HI is stronger than HF (greater size of I courses greater length of bond)
galina
HI is a stronger acid due to less efficient orbital overlap. HF will react with with glass and extract calcium from bones, but those hazards are not because it's a stronger acid, but because it contains fluorine.
Eszter
hi
Victoria
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Jovial
hello
Asha
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Andre
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MO
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nanmya
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they are noble gases