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C4H6Os Malic Acid Tartaric Acid 134 150 9. Calculate the % error between the mol

ID: 1026398 • Letter: C

Question

C4H6Os Malic Acid Tartaric Acid 134 150 9. Calculate the % error between the molar mass of the selected acid and the experimental value. 10. Calculate the % error between the pKas of the selected acid and each experimental value. Describe the confidence in your acid identification based on your answers to this and the preceding question. 11. Consider the titration of 25.00 mL of 0.200 M methyl amine (CH)NH2). The titrant is 0.120 M Hcl. Calculate each of the following: a. the initial pH of the base. b. the pH at 5.00 mL added acid C. the pH at ½ the HCI needed to reach the equivalence point d. the volume of added acid required to reach the equivalence point. e. the pH at the equivalence point

Explanation / Answer

methyl amine= 25.00 ml of 0.200M

number of moles of methyl amine= 0.200M x0.025 L= 0.005 moles

Kb of CH3NH2 = 4.4x10^-4

a)

for weak bases

[OH-]= sqaure root of KbxC

[OH-] = square root of (4.4x10^-4x0.200)

[OH-] = 0.938 x10^-2M

-log[OH-] = -log[0.938x10^-2]

POH= 2.03

PH+POH= 14

PH= 14 -POH

PH= 14-2.03

PH= 11.97.

b) at 5.00 ml of HCl

HCl= 5.00ml of 0.120M

number of moles of HCl = 0.120Mx0.005L= 0.0006 moles

CH3NH2 + HCl ---------- CH3NH3+Cl-

0.005           0.0006           0

-0.0006      -0.0006           +0.0006

after addition of acid

remaining number of moles of base = 0.005 - 0.0006= -0.0044 moles

number of moles of salt= CH3NH3+Cl- = 0.0006 moles

Kb= 4.4x10^-4

-log(Kb) = -log(4,4x10^-4)

PKb= 3.36

POH= Pkb + log[salt]/[base]

POH= 3.36 + log(0.0006/0.0044)

POH=2.49

PH= 14-2.49 =11.51

PH= 11.51

c) at half equivalent point

POH= PKb

POH= 3.36

PH= 14-3.36=10.64

PH= 10.64

d)

at the equivalent point volume of acid = 0.200x25.0/0.120 = 41.67 ml

volume of HCl at equivalent point = 41,67 ml

e)

at equivalent point

PH= 7- 1/2[PKb + logC]

C= concentration of salt

number o fmoles of base or salt= 0.005 moles

total volume = 25.0 + 41.67= 66.67 ml = 0.06667 L

C= 0.005/0.06667 = 0.07499M

PH= 7 - 1/2[3.36 + log(0.07499)]

PH=5.88

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