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 pointExplanation / 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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