Question

Write complete balanced equations in aqueous solution for the reaction of sulfuric acid with sodium hydroxide and the reaction of acetic acid with potassium hydroxide What is the difference in a standard solution and the analyte? If 46.53 mL of 1.500 M NaOH are required to titrate 25.00 mL of an unknown acetic acid solution (vinegar). a.) how many moles of NaOH were required? Show work b.) How many moles are acetic acid are neutralized? c.) what is the molarity of the acetic acid? Show work. The H* ion concentrations in apricot, asparagus, and banana solutions are approximately 1.6 x 104 M, 2.5 x 10° M, and 2.5 x 10-5 M respectively. Determine the pH of apricot, asparagus, and banana solutions. List the three foods of the previous question in order of decreasing acidity. 50

Answer 1

H_2 SO_4 + 2 NaOH rightarrow Na_2 SO_4 + 2 H_2 O. CH_3 COOH + KOH rightarrow CH_3 COOK + H_2 O Standard solution: The concentration of the solute is known Analyte: The concentration is unknown v_1 = Volume of NaOH = 46.53mL S_1 = 1.500 M NaOH V_2 Volume of unknown acetic acid = 25 mL S_2 = conc. of Acetic acid Formula: V_1 S_1 = V_2 S_2 S_2 = v_1 S_1/V_2 = 46.53 times 1.5/25 M S_2 = 2.7918 M leftarrow Concentration of Acetic acid Moles of NaOH = molarity/1000 times 46.53 moL = 1.5/1000 times 46.53 moL = 0.06979 mol