A 50.0 mL sample of 0.10 M ethanolamine, C2H4(OH)NH2, is titrated with 0.2 M HI. Calculate the pH to one decimal place when the following volumes of titrant have been added. A 50.0 mL sample of 0.10 M ethanolamine, C2H4(OH)NH2, is titrated with 0.2 M HI. Calculate the pH to one decimal place when the following volumes of titrant have been added 0.00mL 19.0mL 25.0mL 34.0ml
SHOW WORK PLEASE
Ethanolamine = C2H4(OH)NH2 = 50.0ml of 0.10M
Kb of Ethanolamine = 3.2 x 10^-5
a) 0.00ml of addition
for weak bases
[OH-] = square root of KbXC
[OH-] = square root of (3.2x10^-5 x 0.10)
[OH-] = 1.789 x10^-3
-loh[OH-] = -log(1.789 x10^-3)
POH = 2.75
PH + POH = 14
PH = 14 - POH
PH = 14 -2.75
PH = 11.25
b) addtion of 19.0 ml of HI
HI= 19.0 mL of 0.2M
number of moles of HI = 0.2M x 0.019L= 0.0038 moles
Ethanolamine = C2H4(OH)NH2 = 50.0ml of 0.10M
number of moles of Ethanolamine = C2H4(OH)NH2 = 0.10M x 0.050L = 0.0050 moles
Kb = 3.2x10^-5
-log(Kb) = -log(3.2x10^-5)
PKb = 4.49
C2H4(OH)NH2 + HI --------------------------- C2H4(OH)NH3+I-
Initial 0.0050 0.0038 0
change - 0.0038 - 0.0038 +0.0038
equilibrium 0.0012 0 +0.0038
POH = PKb + log[salt]/[base]
POH = 4.49 + log(0.0038/0.0012)
POH = 4.99
PH = 14 - 4.99 = 9.01
PH = 9.01
C) addition of 25 mL of HI
HI = 25 mL of 0.2M
number of moles of HI= 0.2M x 0.025L = 0.0050 moles
number of moles of Ethanolamine = 0.0050 moles
It is equivalent point because number of moles of base is equal to acid
at equivalent point
PH = 7 - 1/2 [PKb +logC]
total volume= 50.0mL+ 25mL = 75mL = 0.075L
C= number of moles/volume = 0.0050/0.075 = 0.0667 M
PH = 7 - 1/2 [ 4.49 + log(0.0667)]
PH = 5.34
d) additon of 34 mL of HI
HI = 34mL of 0.2M
number of moles of HI = 0.2M x 0.034L = 0.0068 moles
number of moles of base = 0.0050 moles
number of moles of acid is more than that of base
so the nature of the solution is acidic.
remaining number of moles of acid = 0.0068 - 0.0050 = 0.0018 moles
total volume = 50.0+34.0 = 84.0 mL = 0.084 L
[H+] = 0.0018/0.084 = 0.0214
[H+] = 0.0214M
-log[H+] = -log(0.0214)
PH = 1.67
A 50.0 mL sample of 0.10 M ethanolamine, C2H4(OH)NH2, is titrated with 0.2 M HI. Calculate...
A 50.0 mL sample of 0.16 M aninline, C6H5NH2, is titrated with 0.32 M HBr. Calculate the pH to one decimal place when the following volumes of titrant have been added 0.00 mL 「 14.0 mL 25.0 mL 34.0 mL
A 50.0 mL sample of 0.11 M nitrous acid is titrated with 0.22 M NaOH. Calculate the pH to one decimal place when the following volumes of NaOH have been added. 0.00 mL 22.0 mL 25.0 mL 33.0 mL
A 25.00 mL sample of 0.280 M LiOH is titrated with 0.750 M HI at 25 °C. Calculate the initial pH before any titrant is added. pH = Calculate the pH of the solution after 5.00 mL of the titrant is added. pH =
A 25.00 mL sample of 0.300 M KOH analyte was titrated with 0.750 M HI at 25 °C. Calculate the initial pH before any titrant was added. Number pH= Calculate the pH of the solution after 5.00 mL of the titrant was added. Number pH=
A 25.00 mL sample of 0.280 M LiOH is titrated with 0.750 M HI at 25 °C. initial ph = 13.45 calculate the ph after addition of 5.00 ml titrant is added.
50.0 mL of 0.10 M HNO2 is being titrated with 0.20 M NaOH. What is the pH after 25.0 mL NaOH has been added? What is the pH after 35.0 mL NaOH has been added?
please solve and show work! A 100.0 mL sample of 0.10 M Ba(OH)2 is titrated with 0.10 M HBr. Determine the pH of the solution after the addition of 100.0 mL of 0.10 M HBr. Assume volumes can be added. Hint: Remember 1 mole of Ba(OH)2 will give 2 moles of OH A. 12.00 B. 1.30 C. 12.70 D. 2.00 E. 7.00
A 25.0 ml sample of 0.20 M Formic Acid (HCO2H, aq) is titrated with 0.10 M KOH(aq). What is the pH after 50.0 mL of the 0.10 M KOH has been added?
A 50.0 ml sample of 0.50 M acetic acid, ch3cooh is titrated with a 0.150 M NaOH solution. calculate the ph after 25.0 ml of the base have been added (ka=1.8x10^-5)
Item 9 A 50.0 mL sample of 0.120 M HBr is titrated with 0.240 MNaOH. Calculate the pH after the addition of the following volumes of base. Part A.) 0.0 mL Express your answer using two decimal places. Part B.) 20.0 mL Express your answer using two decimal places. Part C.) 24.9 mL Express your answer using one decimal place. Part D.) 25.0 mL Express your answer using two decimal places. Part E.) 25.1 mL Express your answer using two...