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Acetic acid has a Ka of 1.8*10^-5. Three acetic acid/ acetate buffer solutions, A,B, and C, wer made using varying conce...

Acetic acid has a Ka of 1.8*10^-5. Three acetic acid/ acetate buffer solutions, A,B, and C, wer made using varying concentrations: 1. [acetic acid] ten times greater than [acetate] 2. [acetate] ten times greater than [acetic acid] 3. [acetate] = [acetic acid] Match each buffer to the expected pH pH = 3.74 pH= 4.74 pH = 5.74
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Concepts and reason

The concept used to solve this question is to match the pH{\rm{pH}} with the corresponding buffer solutions A, B and C.

Fundamentals

pHpH is the measure of hydrogen ion concentration. Lower the pHpH , higher is the hydrogen ion concentration and lower is the hydroxide ion concentration.

An equilibrium constant for the dissociation reaction is the equation expressing the extent of dissociation into ions which is equal to the product of the concentrations of the respective ions divided by the concentration of the undissociated molecule.

Consider a dissociation reaction of acetic acid, CH3COOH+H2OH3O++CH3COO{\rm{C}}{{\rm{H}}_3}{\rm{COOH}} + {{\rm{H}}_2}{\rm{O}}\longrightarrow{{}}{{\rm{H}}_3}{{\rm{O}}^ + } + {\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ - } .

The equilibrium constant for the dissociation of HA{\rm{HA}} is given as follows:

Ka=[H3O+][CH3COO][CH3COOH]{K_a} = \frac{{\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ - }} \right]}}{{\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}} \right]}}

pKap{K_a} is also the measure of acidic strength.

The formula relating Ka{K_a} and pKap{K_a} is as follows:

pKa=logKap{K_a} = - \log {K_a}

The Henderson-Hasselbalch equation to calculate the pH{\rm{pH}} of a buffer solution is as follows:

pH=pKa+log[CH3COO][CH3COOH]{\rm{pH}} = p{K_a} + \log \frac{{\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{CO}}{{\rm{O}}^ - }} \right]}}{{\left[ {{\rm{C}}{{\rm{H}}_3}{\rm{COOH}}} \right]}}

Given, Ka=1.8×105{K_a} = 1.8 \times {10^{ - 5}}

The pKap{K_a} of the acid is calculated as follows:

pKa=log(1.8×105)=4.74\begin{array}{l}\\p{K_a} = - \log \left( {1.8 \times {{10}^{ - 5}}} \right)\\\\{\rm{ = 4}}{\rm{.74}}\\\end{array}

Given that, for buffer A,

[Aceticacid]\left[ {{\rm{Acetic acid}}} \right] is 10 times greater the concentration of [Acetate]\left[ {{\rm{Acetate}}} \right] .

So,

[Aceticacid]=10x[Acetate]=x\begin{array}{l}\\\left[ {{\rm{Acetic acid}}} \right] = 10x\\\\\left[ {{\rm{Acetate}}} \right] = x\\\end{array}

Substitute the values in the Henderson-Hasselbalch equation and calculate the pH{\rm{pH}} of the buffer as follows:

pH=4.74+logx10x=3.74\begin{array}{l}\\{\rm{pH}} = 4.74 + \log \frac{x}{{{\rm{10}}x}}\\\\{\rm{ = 3}}{\rm{.74}}\\\end{array}

Given that, for buffer B,

[Acetate]\left[ {{\rm{Acetate}}} \right] is 10 times greater the concentration of [Aceticacid]\left[ {{\rm{Acetic acid}}} \right] .

So,

[Aceticacid]=x[Acetate]=10x\begin{array}{l}\\\left[ {{\rm{Acetic acid}}} \right] = x\\\\\left[ {{\rm{Acetate}}} \right] = 10x\\\end{array}

Substitute the values in the Henderson-Hasselbalch equation and calculate the pH{\rm{pH}} of the buffer as follows:

pH=4.74+log10xx=5.74\begin{array}{l}\\{\rm{pH}} = 4.74 + \log \frac{{10x}}{x}\\\\{\rm{ = 5}}{\rm{.74}}\\\end{array}

Given that, for buffer C,

[Aceticacid]=[Acetate]\left[ {{\rm{Acetic acid}}} \right] = \left[ {{\rm{Acetate}}} \right] .

So,

[Aceticacid]=x[Acetate]=x\begin{array}{l}\\\left[ {{\rm{Acetic acid}}} \right] = x\\\\\left[ {{\rm{Acetate}}} \right] = x\\\end{array}

Substitute the values in the Henderson-Hasselbalch equation and calculate the pH{\rm{pH}} of the buffer as follows:

pH=4.74+logxx=4.74\begin{array}{l}\\{\rm{pH}} = 4.74 + \log \frac{x}{x}\\\\{\rm{ = 4}}{\rm{.74}}\\\end{array}

Ans:

The pH{\rm{pH}} of the three buffer solutions are as follows:

BufferA:3.74BufferB:5.74BufferC:4.74\begin{array}{l}\\{\rm{Buffer A}}:3.74\\\\{\rm{Buffer B}}:5.74\\\\{\rm{Buffer C}}:4.74\\\end{array}

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