Question

**A 35.00−mL solution of 0.2500** *M*
**HF is titrated with a standardized 0.1606**
*M* **solution of NaOH at 25**

°

**C.**

**(a) What is the pH of the HF solution before titrant is
added?**

**(b) How many milliliters of titrant are required to reach
the equivalence point?**

**mL**

**(c) What is the pH at 0.50 mL before the equivalence
point?**

**(d) What is the pH at the equivalence point?**

**(e) What is the pH at 0.50 mL after the equivalence
point?**

Answer #1

HF , Ka = 6.3 x 10^-4

pKa = 3.20

a )

pH = 1/2 (pKa - log C)

= 1/2 (3.20 - log 0.25)

**pH = 1.91**

b)

At equivalence point :

millimoles of acid = millimoles of base

35 x 0.250 = V x 0.1606

V = 54.48 mL

**volume of base = 54.48 mL**

c)

millimoles of acid = 35 x 0.25 = 8.75

millimoles of base = 53.98 x 0.1606 = 8.669

HF + NaOH ----------------> NaF + H2O

8.75 8.669 0 0

0.081 0 8.669

pH = pKa + log [NaF]/[HF]

pH = 3.20 + log (8.669 / 0.081)

**pH = 5.23**

d)

at equivalence point :

salt remains.

salt concentration = 8.75 / 35 + 54.48 = 0.0978 M

pH = 7 + 1/2 (pKa + log C)

= 7 + 1/2 (3.20 + log 0.0978)

**pH = 8.10**

e )

pOH = 3.05

**pH = 10.95**

Consider the titration of 40.0 mL of 0.250 M HF with 0.200 M
NaOH. How many milliliters of base are required to reach the
equivalence point? Calculate the pH at the following points:
a)After the addition of 10.0 mL of base
b)Halfway to the equivalence point
c)At the equivalence point
d)After the addition of 80.0 mL of base

Question1: A 100.0 mL solution of 0.5
M histidine in its tribasic form, A3-, (pKa1
= 1.70, pKa2 = 6.02, pKa3 = 9.08) is titrated
with 1.0 M HCl titrant.
1)calculate the volume of titrant required to reach the second
equivalence point,
2)calculate the pH after the addition of 50.0 mL of titrant,
3)calculate the pH after the addition of 62.0 mL of titrant,
4)calculate the pH after the addition of 75.0 mL of titrant.
Question4: A 100 mL volume...

25.000 mL of 0.1114 M propanoic acid solution is titrated with
0.1308 M NaOH. Calculate the pH of titrant mixture at the following
volumes of NaOH added
a) 12.72 mL
b) 21.29 mL (Equivalence point)

1. A 100 mL solution of 0.200 M HF is titrated with 0.100 M
Ba(OH)2. What is the volume of Ba(OH)2 needed
to reach equivalence point?
200 mL
50 mL
100 mL
300 mL
Cannot determine based on the provided information.
2. A 100 mL solution of 0.200 M NH3 is titrated with
0.100 M HCl. What is the volume of HCl needed to reach equivalence
point?
200 mL
100 mL
50 mL
300 mL
Cannot determine based on the...

A 20.00-mL sample of formic acid (HCO2H) is titrated with a
0.100 M solution of NaOH. To reach the endpoint of the titration,
30.00 mL of NaOH solution is required. Ka = 1.8 x
10-4
What is the pH of the solution after the addition of 10.00 mL of
NaOH solution?
What is the pH at the midpoint of the titration?
What is the pH at the equivalence point?

a.) What is the pH at the equivalence point when a HF solution
is titrated with NaOH? Choose from: less than 7, 7, or greater than
7.
b.)Calculate the pH at the equivalence point when a 200. ml
0.050M HF solution that is titrated with 0.040M NaOH solution?
(Ka HF = 7.2 x 10-4)

A 25.00 mL sample of 0.280 M NaOH analyte was titrated with
0.750 M HCl at 25 °C.
Calculate the initial pH before any titrant was added.
Calculate pH after 5.0 ml of titrant was added

A 25 mL aliquot of an HCl solution is titrated with 0.100 M
NaOH. The equivalence point is reached after 21.27 mL of the base
were added. Calculate the concentration of the acid in the original
solution, the pH of the original HCl solution and the original NaOH
solution

1) If 25.2 mL of 0.109 M acid with a pKa of 5.55 is titrated
with 0.102 M NaOH solution, what is the pH of the titration mixture
after 13.7 mL of base solution is added?
2) If 28.8 mL of 0.108 M acid with a pKa 4.15 is titrated with
0.108 M NaOH solution, what is the pH of the acid solution before
any base solution is added?
3)If 27.9 mL of 0.107 M acid with a pKa of...

1. 20.00 mL of 0.510 M NaOH is titrated with 0.740 M
H2SO4. ______ 1 mL of
H2SO4 are needed to reach the end point.
2. If 35.00 mL of 0.0200 M aqueous HCl is required to
titrate 30.00 mL of an aqueous solution of NaOH to the equivalence
point, the molarity of the NaOH solution is ________ 1 M.

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 minute ago

asked 4 minutes ago

asked 12 minutes ago

asked 15 minutes ago

asked 21 minutes ago

asked 24 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago