No addition of [tex]{mathbf{0}}{mathbf{.020 mol}}[/tex] of [tex]{mathbf{NaOH}}[/tex] will not exceed buffer capacity.n

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Further explanation:n

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Buffer capacity refers to the amount of acid or base that a buffer can neutralizebefore its pH changes appreciably. In general, the maximum buffer capacity exists when the concentrations of a weak acid and its conjugate base are kept large and approximately equal to each other. The buffer range is the pH range over which a buffer effectively neutralizes added acids and [tex]0.020{text{ mol}}[/tex] of [tex]{text{NaOH}}[/tex] and maintains a fairly constant .

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As henderson-hasselbalch equation suggests when the ratio falls to 0.10, then decreases by 1 unit. If the ratio increases to a value of 10, then increase by 1 unit.n

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The ionization of a hypothetical weak acid [tex]left( {{text{HA}}} right)[/tex] is given as follows:n

If one starts with [tex]0.10{text{ mol HA}}[/tex] and [tex]0.2{:text{mol:}}{{text{A}}^ – }[/tex] , then on adding [tex]0.020{text{ mol }}{{text{A}}^ – }[/tex] the concentration of [tex]{{text{A}}^ – }[/tex] increases by this amount while the concentration of [tex]{{text{A}}^ – }[/tex] decreases by [tex]0.020{text{ mol}}[/tex] and the concentration of [tex]{{text{A}}^ – }[/tex] increases by [tex]0.020{text{ mol}}[/tex]. Therefore the net concentration of acid that remains at equilibrium is given as follows:n

The ratio of [tex]dfrac{{left[ {{{text{A}}^ – }} right]}}{{left[ {{text{HA}}} right]}}[/tex] is 2.75. Hence as there is no appreciable fall in this ratio so the buffer capacity is not exceeded upon addition of [tex]{mathbf{0}}{mathbf{.020 mol}}[/tex] of [tex]{mathbf{NaOH}}[/tex].n

## Answers ( 2 )

A buffer solution is a solution whose pH does not vary or little when adding an acid or a base, or during a dilution.

The buffering capacity is the number of moles of acid or strong base to add to 1 L of buffer solution to vary the pH of one unit.

So if we add 0.020 mol of NaOH this will vary the dpH depending on the buffer capacity of the solution.

We Have : Cacid (AH) = 0.1mol/l

Cbase (A-) = 0.2mol/l

The reaction of the buffer with NaOH will be:

AH + NaOH = A- + Na + H2O

t0 0.1mol 0.02mol 0.2mol

t1 0.08mol 0 0.22mol

At the end of the reaction we will have:

Concentration of acid = 0.08mol/lConcentration of the conjugate = 0.22mol/lSo the capacity of the buffer will not be exceed, it still remains both parts of the buffer

No addition of [tex]{mathbf{0}}{mathbf{.020 mol}}[/tex] of [tex]{mathbf{NaOH}}[/tex] will not exceed buffer capacity.nnFurther explanation:nn

Buffer capacityrefers to the amount of acid or base that a buffer can neutralizebefore itspHchanges appreciably. In general, the maximum buffer capacity exists when the concentrations of a weak acid and itsconjugate baseare kept large and approximately equal to each other. The buffer range is the pH range over which a buffer effectively neutralizes added acids and [tex]0.020{text{ mol}}[/tex] of [tex]{text{NaOH}}[/tex] and maintains a fairly constant .n

As

henderson-hasselbalch equationsuggests when theratiofalls to 0.10, then decreases by 1 unit. If the ratio increases to a value of 10, then increase by 1 unit.nn

The ionization of a hypothetical

weak acid[tex]left( {{text{HA}}} right)[/tex] is given as follows:nn[tex]{text{HA}}left( {{text{aq}}} right) + {text{O}}{{text{H}}^ – }left( {text{l}} right) rightleftharpoons {{text{A}}^ – }left( {{text{aq}}} right) + {{text{H}}_3}{{text{O}}^ + }left( {{text{aq}}} right)[/tex]

If one starts with [tex]0.10{text{ mol HA}}[/tex] and [tex]0.2{:text{mol:}}{{text{A}}^ – }[/tex] , then on adding [tex]0.020{text{ mol }}{{text{A}}^ – }[/tex] the concentration of [tex]{{text{A}}^ – }[/tex] increases by this amount while the concentration of [tex]{{text{A}}^ – }[/tex] decreases by [tex]0.020{text{ mol}}[/tex] and the concentration of [tex]{{text{A}}^ – }[/tex] increases by [tex]0.020{text{ mol}}[/tex]. Therefore the net concentration of acid that remains at equilibrium is given as follows:n

[tex]begin{aligned}left[ {{text{HA}}} right] &= 0.10{text{ mol}} – 0.020{text{ mol}}&= {text{0}}{text{.08 mol}}end{aligned}[/tex]

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The ratio of [tex]dfrac{{left[ {{{text{A}}^- }}right]}}{{left[{{text{HA}}} right]}}[/tex] is calculated as follows:n

[tex]begin{aligned}frac{{left[ {{{text{A}}^ – }} right]}}{{left[ {{text{HA}}} right]}} &= frac{{0.22}}{{0.08}}&= 2.75end{aligned}[/tex]

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The ratio of [tex]dfrac{{left[ {{{text{A}}^ – }} right]}}{{left[ {{text{HA}}} right]}}[/tex] is 2.75.

Hence as there is no appreciable fall in this ratio so the buffer capacity is not exceeded upon addition of [tex]{mathbf{0}}{mathbf{.020 mol}}[/tex] of [tex]{mathbf{NaOH}}[/tex].nnLearn more:n1.Balanced chemical equation: https://brainly.com/question/1405182n2.Number of covalent bonds does nitrogen formed with its unpaired: https://brainly.com/question/5974553nn

Answer details:Grade:CollegenSubject:ChemistrynChapter:Molecular structure and chemical bondingnn

Keywords:Buffer capacity, buffer, pH, Henderson-HasselBalch equation, ratio, and HA.n