php/ 20 %20Extra%20Credit%20-%20Spring%2020 18.pdf CHEM 108- Extra Credit-Spring

Question

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?

Explanation / Answer

(5)

(a)

Initial [OH-] = (Kb*C)0.5 = (0.175*4.38*10-4)0.5 = 8.75*10-3 M

Initial pOH = 3-log(8.75) = 2.057

Initial pH = 14-2.057 = 11.943

(b)

Using relation:

M1V1 = M2V2

Putting values:

25*0.175 = 0.15*V2

Solving we get:

V2 = 29.17 mL

(c)

At equivalence point, [CH3NH3+] = (0.025*0.175)/(0.025+0.02917) = 0.0807 M

So, at this point, [H+] = (Ka*C)0.5 = (0.0807*((10-14)/(4.38*10-4)))0.5 = 1.35*10-6 M

So,

pH = 6-log(1.35) = 5.87

(d)

Using Henderson Hasselbach equation:

pOH = pKb + log(moles of salt/moles of base)

Putting values:

pOH = 3.36 + log((0.005*0.15)/(0.025*0.175 - 0.005*0.15)) = 2.675

So,

pH = 14-2.675 = 11.325

(e)

Excess acid added after reaching equivalence point = 35-29.17 = 5.83 mL

So, [H+] due to excess acid = (0.150*0.00583)/(0.035+0.025) = 0.0146 M

So,

pH = -log(0.0146) = 1.835

Hope this helps !

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