Answer to Question #97413 in General Chemistry for Boye Zephaniah

Question

(a) describe the preparation of 500mL of 0.08M acetate buffer (pH 5.4) from acetic acid [ F.W.=60.05, specific gravity= 1.05, 99.6%w/w] and solid NaOH (F.W=40.0).
(b) what is the concentration of Na+ in the buffer?

Answer 1

The pH value of the acetate buffer is governed with the following expression:

$pH=pK_{HA}+lg(\frac{C_{BA}}{C_{HA}})$;

HA is acetic acid and BA is sodium acetate.

From this expression we can found the relation between sodium acetate concentration and acetic acid concentration, which is required for the preparation of the buffer solution:

$\frac{C_{BA}}{C_{HA}}=10^{pH-pK_{HA}}=10^{5.40-4.75}=4.47;$

Now, the concentration of sodium acetate in the buffer should be 0.08M and the concentration of free acetic acid should be:

$C_{HA}=\frac{C_{BA}}{4.47}=\frac{0.08M}{4.47}=0.018M;$

For preparation of the buffer we should mix sodium hydroxide and pure acetic acid:

$CH_3COOH+NaOH \rightarrow CH_3COONa+H_2O$;

The reactants react in a 1 to 1 proportion. Thus, the amount in moles of NaOH required will be the same as the amount of CH₃COONa produced. However, the amount of CH₃COOH required will be the sum of the acid consumed in the reaction with NaOH and the free acid in the solution:

$n(NaOH)=n(CH_3COOH)=C_{BA}*V_{buf}=0.08M*0.5L=0.04mol;$

$m(NaOH)=n(NaOH)*M(NaOH)=0.04mol*40.0g/mol=1.6g;$

$n(CH_3COOH)=n(NaOH)+C_{HA}*V_{buf}=0.04mol+0.018M*0.5L=0.049mol;$

$m(CH_3COOH)=n(CH_3COOH)*M(CH_3COOH)=0.049mol*60.05g/mol=2.9g;$

$m_{sol}(CH_3COOH)=\frac{m(CH_3COOH)*100\%}{w_{sol}}=\frac{2.9g*100\%}{99.6\%}=2.91g;$

$V_{sol}(CH_3COOH)=\frac{m_{sol}(CH_3COOH)}{d_{sol}}=\frac{2.91g}{1.05g/mL}=2.77mL$;

Thus, we should mix 1.6g of a solid NaOH and 2.77mL of a 99.6% liquid acetic acid in an analytical 0.5L flask and dilute everything with some distilled water up to the line on the neck of the flask.