Chemistry, 25.07.2019 16:00 austinbeesley9855

Determine whether or not adding 0.020 mol of naoh exceed the capacity of the buffer.

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Answer from: austinlogan3218

No addition of ${\mathbf{0}}{\mathbf{.020 mol}}$ of ${\mathbf{NaOH}}$ will not exceed buffer capacity.

Further explanation:

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 $0.020{\text{ mol}}$ of ${\text{NaOH}}$ and maintains a fairly constant .

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.

The ionization of a hypothetical weak acid $\left( {{\text{HA}}} \right)$ is given as follows:

${\text{HA}}\left( {{\text{aq}}} \right) + {\text{O}}{{\text{H}}^ - }\left( {\text{l}} \right) \rightleftharpoons {{\text{A}}^ - }\left( {{\text{aq}}} \right) + {{\text{H}}_3}{{\text{O}}^ + }\left( {{\text{aq}}} \right)$

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

$\begin{aligned}\left[ {{\text{HA}}} \right] &= 0.10{\text{ mol}} - 0.020{\text{ mol}}\\&= {\text{0}}{\text{.08 mol}}\\\end{aligned}$

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

$\begin{aligned}\frac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}} &= \frac{{0.22}}{{0.08}}\\&= 2.75\\\end{aligned}$

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

Learn more:

1. Balanced chemical equation:

2. Number of covalent bonds does nitrogen formed with its unpaired:

Answer details:

Grade: College

Subject: Chemistry

Chapter: Molecular structure and chemical bonding

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

Answer from: Quest

3 1/4 %

explanation:

Answer from: Quest

answer: $2c_5h_5+1fe\rightarrow 1fe(c_5h_5)_2$

the reaction is type of combination reaction.

explanation:

$? c_5h_5+? fe\rightarrow ? fe(c_5h_5)_2$

on the product side, there are 2 molecules of $c_5h_5$ bonded to fe which means one atom of fe are combined two molecules of $c_5h_5$ . so, balanced chemical equation becomes :

$2c_5h_5+1fe\rightarrow 1fe(c_5h_5)_2$

this reaction follows the reaction type of combination reaction. a combination reaction is defined as chemical reaction in which reactants combines together to form a single product.

$a+b\rightarrow ab$

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Determine whether or not adding 0.020 mol of naoh exceed the capacity of the buffer....

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