Question

Consider titration of 100.0 mL of 0.100 M H2NNH2 (Kb = 3.5x10^-6) by 0.200 M HNO3? Calculate the pH of the resulting solution after the following volumes of HNO3 have been added.

0.0 mL

20.0 mL

25.0 mL

40.0 mL

50.0 mL

100.0 ml

Can anyone answer this thouroughly please? Please be sure to show every step of the math, that us were I tend to get lost. Thank you!

Answer #1

a) 0.0 mL

pOH = 1/2 [pKb - logC]

pOH = 1/2 [5.46 - log 0.1]

pOH = 3.23

pH + pOH = 14

**pH = 10.77**

b) 20 mL

millimoles of B = 100 x 0.1 = 10

millimoles of acid = 20 x 0.2 = 4

millimoles of B = 10-4 = 6

pOH = pKb + log [4 / 6]

pOH = 5.46 + log [4 / 6]

pOH = 5.28

**pH = 8.72**

c) 25 mL

it is half equivalence point ; here pOH = pKb

pOH = 5.46

**pH =8.54**

d ) 40 mL

millimoles of acid = 40 x 0.2 = 8

B remains = 10-8 = 2

POH = 5.46 + log (8 / 2)

pOH = 6.06

**pH = 7.94**

.Consider the titration of 100.0 mL of 0.100 M H2NNH2
(Kb=3.0E-6) by 0.200 M HNO3. Calculate the pH of the resulting
solution after the following volumes of HNO3 have been added.
A) 0.0 mL B) 20.0 mL C) 25.0 mL D) 40.0 mL E) 50.0 mL F)100.0
mL

Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M
HCl. For each volume of HCl added, decide which of the components
is a major species after the HCl has reacted completely. Kb for
CH3NH2 = 4.4 x 10-4. 1) Calculate the pH at the equivalence point
for this titration 2) At what volume of HCl added in this titration
does the pH = 10.64? Express your answer in mL and include the
units in your answer.

Consider the titration of 50.0 ml of 0.100 M methylamine (Kb =
5.0 × 10- 4) with 0.200 M HCl. What is the pH of the solution when
25.00 ml of HCL has been added?

A 50.0 mL sample of 0.150 M ammonia (NH3, Kb=1.8×10−5) is
titrated with 0.150 M HNO3. Calculate the pH after the addition of
each of the following volumes of acid.
A. 0.0 mL
B. 25.0 mL
C. 50.0 mL

1. A25.0 mL sample of 0.100 M lactic acid (hc3h5o5, pka= 3.86)
is titrated with a 0.100 M NaOH solution. Calculate the pH after
the addition of 0.0, 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9, 25.0,
25.1, 26.0, 28.0, and 30.0 of the NaOH. Plot the results of your
calculation, as a pH versus mililiters of NaOH added.
2. Repeat the procedure in problem 1, but the titration of 25.0
mL 0.100 M NH3 (kb= 1.8*10^-5) with 0.100 M HCl....

Consider the titration of 30.0 mL of 0.0700 M NH3 (a weak base;
Kb = 1.80e-05) with 0.100 M HBrO4. Calculate the pH after the
following volumes of titrant have been added:
(a) 0.0 mL pH =
(b) 5.3 mL pH =
(c) 10.5 mL pH =
(d) 15.8 mL pH =
(e) 21.0 mL pH =
(f) 35.7 mL pH =

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with
0.500 M HNO3. Calculate the pH after the addition of 25.0 mL of
HNO3.

A 102.0 mL sample of 0.100 M methylamine (CH3NH2;Kb=3.7×10−4) is
titrated with 0.270 M HNO3. Calculate the pH after the addition of
each of the following volumes of acid:
Part A: 0.0 mL
Part B: 18.9 mL
Part C: 37.8 mL
Part D: 56.7 mL

A 111.8 mL sample of 0.100 M methylamine
(CH3NH2;Kb=3.7×10−4) is titrated with 0.245
M HNO3. Calculate the pH after the addition of
each of the following volumes of acid.
Part A
0.0 mL
Part B
22.8 mL
Part C
45.6 mL
Part D
68.4 mL

a.) Calculate the pH during the titration of 100.0
mL of 0.200 M HCl with 0.400 M NaOH. First what is the initial pH
(before any NaOH is added)?
b.) What is the pH after 31.9 mL of NaOH are added?
c.) What is the pH after 50 mL of NaOH are added?
d.) What is the pH after 68.9 mL of NaOH are added?

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