Question

1. 20.00 mL of 0.510 M NaOH is titrated with 0.740 M
H_{2}SO_{4}. ______ 1 mL of
H_{2}SO_{4} 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.

Answer #1

Part A

first write the balanced equation

H2SO4 + 2NaOH -------> Na2SO4 + 2H2O

from the balanced equation it is clear that one mole of H2SO4 required 2 moles of NaOH

no of moles of NaOH = molarity x volume of NaOH in liters = 0.51 x 0.02 = 0.0102 moles

no of moles of H2SO4 required for 0.0102 moles of NaOH = 0.0102 /2 = 0.0051 moles

now we have moles of H2SO4 = 0.0051 and molarity of H2SO4 = 0.74 M

Volume of HCl = no of moles / Molarity = 0.0051 / 0.74 = 0.007L

= 0.007 x 1000 ml

= **7 ml of H2SO4**

Part-2

first write the balanced equation

NaOH + HCl ------> NaCl + H2O

no o fmoles of HCl = molarity x volume in liters = 0.02M x 0.035L = 0.0007 moles

from the balanced equation one mole oh HCl required one mole NaOH

so no of moles of NaOH = 0.0007, volume of NaOH given = 30.0 mL = 0.03L

Molarity = moles/liters = 0.0007 mol / 0.03 L

= 0.023M

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?

1.Write the molecular equation and the net ionic equation for
the reaction of H2SO4 with NaOH.
2. Exactly 25.00 mL of 0.1522 M
H2SO4 were needed to titrate 45.25 mL of NaOH according to the
balanced equation in the problem above. Calculate the moles of NaOH
needed for the reaction. Calculate the molarity of the NaOH
solution.
3. One group of students made an
error by using HCl instead of H2SO4 to titrate an unknown solution
of NaOH. Would this...

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 concentration of formic acid in the original
solution?
What is the pH of the formic acid solution before the titration
begins (before the addition of any NaOH)?

The sample of 30.00 mL of 0.015 M HClO is titrated with a 20.00
mL of a 0.015 M NaOH solution.
(a) Show the reaction between HClO and
OH–. (1 point)
(b) What is the final volume after two
solutions were mixed? (1 point)
(c) What is the final molarity of HClO after
the reaction with OH–? (1 point)
(d) What is the pH after 20.0 mL of the NaOH
solution is added? (Ka of HClO is
3.0*10–8) (1
point)

A
25.00 mL sample of an H2SO4 solution of unknown concentration is
titrated with a 0.1322 M KOH solution. A volume of 41.22 mL of KOH
is required to reach the equivalence point. What is the
concentration of the unknown H2SO4 solution? Express your answer in
molarity to four significant figures.

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...

14.27mL of a solution of the diprotic acid
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A 10.00 mL solution of 0.1900 M NH4Cl is titrated
with 0.1300 M NaOH to the equivalence point.
Volume of NaOH = 14.62mL
Calculate the pH of the solution at the equivalence
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Calculate the pH of a simple 0.0772 M aqueous solution
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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
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(d) What is the pH at the equivalence point?
(e) What is the pH at 0.50 mL after the equivalence...

A 100 mL solution of 0.200 M Sr(OH)2 is titrated with
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100 mL
200 mL
50 mL
300 mL
133 mL
A 100 mL solution of unknown concentration
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