Question

Please answer #2

php/ 20 %20Extra%20Credit%20-%20Spring%2020 18.pdf CHEM 108- Extra Credit-Spring 2018 IMPORTANT: Show ALL your work. Don't forget the significant figures and units!! Would the following mixtures result in buffer solutions? (Justify your answers) 1. a 100.0 ml, of 0.10 M NH, 100.0 mL of0.15 M NHaCl b. 50.0 mL of 0.10 M HCIO4, 35.0 mL if 0.15 M NaCIO c. 125.0 ml, of 0.15 CH?NH2. 120.0 mL of 0.25 M CHNE ICI d. 165.0 mL of 0.10 M HF, 135.0 mL of 0.175 M HCT e. 75.0 mL of 0.10 HF, 55.0 mL. of O.15 NaF Calculate the pH of a buffer solution that results from mixing 60.0 mL ifo.250 M HCOOH and 15.0 mL if 0.500 M NaCOOH using BOTH ICE table AND the Henderson-Hasselbalch equation for comparison. Ka-1.77 x i0" fr HGOOH 2. 3. A 250.0 mL buffer solution is 0.250 M in acetic acid and 0.250 M is sodium acetate a. What is the initial pH of the solution? b. What is the pH after addition of 0.0050 mol of HCT? c. What is the phl after addition of 0.0050 mol of NaOH? 4. 35.0 ml of 0.175 M HBr solution are titrated with a 0.200 M KOH solution a. What is the initial pH? b. What is the volume of KOH required to reach the equivalence? c. What is the pH at the equivalence? d. What is the plH after 10 mL of base are added? e. What is the pH after 45 mL of base are added? 25.0 ml of0.175 M CHiNH; solution (kb-438 x 10) are titrated with a 0.150 M HBr solution: 5. a. What is the initial pH? b. What is the volume of acid required to reach the equivalence? c. What is the pH at the equivalence? d. What is the pll after 5.0 ml of acid are added? e. What is the pH after 35.0 ml. of acid are added?

Answer 1

2 > p_H = ? HCOOH rightleftarpoons H^+ + HCOO^- K_a = 1.77 times 10^-4 NaCOOH rightleftarpoons Na^+ + HCOO^- H is a Mixture OF a Weak Acid and its salt with Strong Base. MilliMoles OF Acid = 60.0 mL times 0.250 M = 15 MilliMoles MilliMoles OF salt = 15.0 mL times 0.500 M = 7.5 MilliMoles Total Volume = 60.0 + 15.0 = 75.0mL Acid rightarrow [HCOOH] = 15/75 = 0.2 M Salt rightarrow = 7.5/75 = 0.1 M According to HENDERSON - HASSELBALCH EQUATION pK_a = -log[K_a] = -log[1.77 times 10^-4] = 3.75 p_H = pK_a + log[Salt]/[Acid] p_H = 3.75 + log 0.1/0.2 p_H = 3.75 + log 0.5 p_H = 3.75 - 0.301 p_H = 3.45

Note: Compare the result using Ice table.