Q1. The molality of 1/10 vinegar is 1.665 M ± 0.1. 10% Vinegar means that it has 10% of the weight of Vinegar is acetic acid and the other 90% is water. Thus, it will be 10g out of 100 of acetic acid which is equal to 10 g / 60.05 g/mol = .1665 moles. The 100g of 10% vinegar would have a volume of 100 ml (0.1 L) which means, the molarity will be the number of moles per liter. (0.1665/0.1= 1.665 M).
Q2. The wt% acetic acid is 5 % which is true since vinegar comprises of acetic acid and water and if the acetic acid is 5%, it means that the solution of vinegar has 5 ml of acetic acid out of the total volume of vinegar.
Q3. The calculated pka for 5% vinegar is similar to that of acetic acid that equals 4.74.
Q4. pka is temperature dependent since an increase in temperature results in a decrease in pka. Ka tends to increase with temperature, and the acid gets stronger than before. However, pka is –log Ka which means that as ka increases, there will be a corresponding decrease in pka.
Using: pKa2-pKa1= ∆H/(R Ln(10) )*(1/T2-1/T1)
∆H=394.439l/mol
R=9.314 J/K. mol
Known pka2=4.756 at 298.15 K
Experimental temperature was 303K.
Thus, by rearranging the formula: pka1=pka2 – ∆H/(R Ln(10) )*(1/T2-1/T1)
Pka1= 4.756- {(394.439/9.314 ln (10)} *(1/298.15-1/303)
Pka 1= 4.755 (Calculated pka1 at 303 K)
Q5. The calculated pka and the derivative are different by a small margin since the two experiments were conducted at different temperatures. The experiment conducted at a lower temperature (298.15 K) had a higher pka than that conducted at a slightly higher temperature (303 K).
Second page
Figure 25a.2. shows the graph of the second derivative of pH against the added NaOH in ml. The equivalence point is 12.61 ml at a temperature of 294.2 K. The point is obtained by identifying the exact half-titration point of an acid solution. From the graph shown, the equivalence point means that at a volume of 12.61 ml of sodium hydroxide added to acetic acid, the PH of the second derivative of PH of the solution is zero.
Figure 25a3 shows the inflection point upon the addition of sodium hydroxide (a base) to an acidic solution and the effect of PH. The PH is expected to increase since the solution becomes less and less acidic as more sodium hydroxide is added. As the additional base is added, it neutralizes the hydrogen ions and the positive log term increases in the pka result. The exact point where the log. The term is zero, and the PH is approximately equal to pka is the inflection point as shown on the graph. Thus, pka is approximately equal to 4.62 when the temperature of the solution is 294.2 K.
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