A student following the procedure in this experiment prepared three solutions by
ID: 506665 • Letter: A
Question
A student following the procedure in this experiment prepared three solutions by adding 11.50, 23.00, and 34.50 mL of a 7.60 times 10^-2 M NaOH solution to 50.00 mL of a 0.112M solution of a weak acid. The solutions were labeled 1, 2, and 3, respectively. Each of the solutions was diluted to a total volume of 100.0 mL with distilled water. The pH readings of these solutions were; (1) 6.42, (2) 6.81, and (3) 7.09. Convert the pH of each solution to an equivalent H_3O^+ ion concentration. For each solution, calculate the number of moles of acid added. For each solution, calculate the number of moles of OH^- ion added. For each solution, determine the number of moles of HAn and of An^- ion at equilibrium.Explanation / Answer
1) We define pH of a solution as
pH = -log [H3O+]
===> [H3O+] = antilog (-pH)
Solution 1: pH = 6.42; [H3O+] = antilog (-6.42) = 3.80*10-7 M (ans)
Solution 2: pH = 6.81; [H3O+] = antilog (-6.81) = 1.549*10-7 M (ans)
Solution 3: pH = 7.09; [H3O+] = antilog (-7.09) = 8.128*10-8 M (ans)
2) 50.00 mL of 0.112 M weak acid was added to all the three solutions. Therefore, the moles of weak acid added to all the three solutions = (50.00 mL)*(1 L/1000 mL)*(0.112 mol/L) = 0.0056 mole (ans).
3) Solution 1: 11.50 mL of 7.60*10-2 M NaOH is added; moles of NaOH added = (11.50 mL)*(1 L/1000 mL)*(7.60*10-2 mol/L) = 8.74*10-4 mole (ans).
Solution 2: 23.00 mL of 7.60*10-2 M NaOH is added; moles of NaOH added = (23.00 mL)*(1 L/1000 mL)*(7.60*10-2 mol/L) = 1.748*10-3 mole (ans).
Solution 3: 34.50 mL of 7.60*10-2 M NaOH is added; moles of NaOH added = (34.50 mL)*(1 L/1000 mL)*(7.60*10-2 mol/L) = 2.622*10-3 mole (ans).
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.