Question

Answer only the questions on the back page (the empty side)

Mandelic Acid Data & Results. Fill in the table and in the space below it show how you calculated the solution's molarity and the acidity constant for one entry solution (mol/L) Measured pH Acidity co Ka Mass of mandelicMolarity of acid (g) 2. LuL l a oz 9? Table 1 of Data and Results 1310 1S1. ? 3. 0L : 0. 01 9 1 51.14 Acetic Acid Data & Results. Fill in the table and in the space below it show how you calculated the solution's molarity and the acidity constant for one entry Volume of Actual Acid used (mL) Molarity (M) Measured pH Acidity constant, pH Ka 2.413 2 30 ‘? a0 m 25 Table 2 of Data and Results

Answer 1

Commercial Vinegar Data and Result: Labled Percent acidity Measured p^H [H^+] 7.6 7.382 4.14 times 10^-8 p^h = -log[H^+] = > - log[H^+] = > [H^+] = Antilog(- 7.382) = 4.14 times 10^-8 Average ka for acetic acid from table 2 Acetic Acid Average ka 4.4255 times 10^-6 3.96 times 10^-6 + 4.891 times 10^-6 /2 = 4.4255 times 10^-6 \r\n Average p^H from the table 2 = 2.397 - log[H^+] = 2.397 = > [H^+] = antilog(- 2.397) = 4.008 times 10^-3 Therefore [H^+] = squareroot e ka = > (4.008 times 10^-3)^2 = e. ka = > e = (4.008 times 10^-3)^2/4.4 times 10^-6 concentration ((0) of acetic acid in commercial Vinegar \r\n calculation for the determination of acetic acid in Vinegar Man of Vinegar used in experiment(say) = 10g Balance equation for the reaction NaOH + CCH_3COOH rightarrow CH_3COONa + H_2O so perform this need 0.1 m or 0.2m NaoH the number of gram of acetic acid by the sodium = 0.2m times 60g(CH_3COOH)(1 mole) times 0.0250L = 0.3 g CH_3COOH % acetic acid = (man of acetic acid /man of Vinegar) times 100 = 0.3g CH_3COOH/10.00 g Vinegar times 100 = 3% acetic acid