Question

Consider the titration of a 25.0 mL sample of 0.110 molL−1
CH3COOH (*K*a=1.8×10−5) with 0.130 molL−1 NaOH. Determine
each quantity:

**Part A**

Part complete

the initial pH

Express your answer using two decimal places.

pH = |
2.85 |

Submit**Previous Answers**

Correct

**Part B**

Part complete

the volume of added base required to reach the equivalence point

V = |
21.2 |
mL |

Submit**Previous Answers**

Correct

**Part C**

Part complete

the pH at 4.0 mL of added base

Express your answer using two decimal places.

pH = |
4.11 |

Submit**Previous Answers**

Correct

**Part D**

Part complete

the pH at one-half of the equivalence point

Express your answer using two decimal places.

pH = |
4.74 |

Submit**Previous Answers**

Correct

**Part E**

Part complete

the pH at the equivalence point

Express your answer using two decimal places.

pH = |
8.76 |

Submit**Previous Answers**

Correct

**Part F**

the pH after adding 5.0 mL of base beyond the equivalence point

Express your answer using two decimal places.

**CAN SOMEONE ANSWER QUESTION F ONLY
PLEASE!!!!!!!!**

Answer #1

F)

Given:

M(CH3COOH) = 0.11 M

V(CH3COOH) = 25 mL

M(NaOH) = 0.13 M

V(NaOH) = 5 mL more than equivalence point

= 5 mL + 21.2 mL

= 26.2 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.11 M * 25 mL = 2.75 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.13 M * 26.2 mL = 3.406 mmol

We have:

mol(CH3COOH) = 2.75 mmol

mol(NaOH) = 3.406 mmol

2.75 mmol of both will react

excess NaOH remaining = 0.656 mmol

Volume of Solution = 25 + 26.2 = 51.2 mL

[OH-] = 0.656 mmol/51.2 mL = 0.0128 M

use:

pOH = -log [OH-]

= -log (1.281*10^-2)

= 1.8924

use:

PH = 14 - pOH

= 14 - 1.8924

= 12.1076

Answer: 12.11

Consider the titration of 39.2 mL of 0.265 M HF with 0.220 M
NaOH. Calculate the pH at each of the following points.
Part A
Calculate the pH after the addition of 9.80 mL of base. Express
your answer using two decimal places.
Part B
Calculate the pH at halfway to the equivalence point. Express your
answer using two decimal places.
Part C
Calculate the pH at the equivalence point. Express your answer
using two decimal places.
Part D
Calculate...

Consider the titration of 35.0 mL of 0.260 M HF with 0.215
MNaOH. Calculate the pH at each of the following points.
Part A.)
How many milliliters of base are required to reach the
equivalence point?
Express your answer using three significant figures.
Part B.)
Calculate the pH after the addition of 8.75 mL of base
Express your answer using two decimal places.
Part C.)
Calculate the pH at halfway to the equivalence point.
Express your answer using two decimal...

Consider the titration of a 25.0 −mL sample of 0.110 M HC7H5O2
with 0.120 M KOH. Determine each of the following.
The Ka of HC7H5O2 is 6.5 × 10−5.
Part A: the pH at 5.00 mL of added base pH= ?

Consider the titration of a 27.8 −mL sample of 0.125 M RbOH with
0.110 M HCl. Determine each of the following. A) the initial pH
B)the volume of added acid required to reach the equivalence point
C) the pH at 6.0 mL of added acid D) the pH at the equivalence
point E) the pH after adding 4.0 mL of acid beyond the equivalence
point

eight
Consider the titration of a 35.0ml sample of 0.175 M HBr with
.205M KOH. Determine the following.
partA the initial pH express answer
using three decimal places
PartB the volume of added base required to
reach the equivalence point express answer in
millimeters
PartC the pH at 11.9mL of added base
express answer using three decimal places
PartD the pH at the equilalence point
express answer as a whole number
PartE the pH after adding 5.0 mL of base...

Consider the titration of a 27.0 −mL sample of 0.175 MCH3NH2
with 0.150 M HBr. Express answers using two decimal places.
Determine each of the following.
A) the initial pH
B) the volume of added acid
required to reach the equivalence point
C) the pH at 6.0 mL of
added acid
D) the pH at one-half of the equivalence point
E) the pH at the equivalence point
F) the pH after adding 4.0 mL of acid beyond the equivalence
point

Consider the titration of 36.4mL of 0.265M HF with 0.220M NaOH.
Calculate the pH at each of the following points.
Part B
Calculate the pH after the addition of 9.10mL of base
Express your answer using two decimal places.
Part C
Calculate the pH at halfway to the equivalence point.
Express your answer using two decimal places.
Part D
Calculate the pH at the equivalence point.
Express your answer using two decimal places.
Part E
Calculate the pH after the...

Consider the titration of a 21.0 −mL sample of 0.100 M HC2H3O2
with 0.125 M NaOH. Determine each of the following.
A. the initial pH = 2.87
B. the volume of added base required to reach the equivalence
point = 16.8 mL
C. the pH at 6.00 mL of added
base = 4.49
D. the pH at one-half of the equivalence point =
4.74
F. the pH after adding
6.00 mL of base beyond the equivalence point.... I need help...

Consider the titration of 51.0 mL of 0.200 MHNO3 with 0.435 M
NaOH.
Part A
How many millimoles of HNO3 are present at the start of the
titration?
Express your answer using three significant figures.
Part B
How many milliliters of NaOH are required to reach the
equivalence point?
Express your answer using three significant figures.
Part C
What is the pH at the equivalence point?
Express your answer using two decimal places.

Problem 16.45 part 2
Part E
Calculate [OH−] for 1.60 mL of 0.130 M NaOH diluted to 1.50 L
.
Express your answer using three significant figures.
[OH−] =
M
SubmitMy AnswersGive
Up
Part F
Calculate pH for 1.60 mL of 0.130 M NaOH diluted to 1.50 L .
Express your answer using three decimal places.
pH =
SubmitMy AnswersGive
Up
Part G
Calculate [OH−] for a solution formed by adding 4.70 mL of 0.150
M KOH to 20.0 mL...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 4 minutes ago

asked 4 minutes ago

asked 13 minutes ago

asked 22 minutes ago

asked 23 minutes ago

asked 27 minutes ago

asked 36 minutes ago

asked 36 minutes ago

asked 43 minutes ago

asked 43 minutes ago

asked 47 minutes ago

asked 51 minutes ago