Question

A 40.0 mL volume of 0.100 M NaOH is titrated with 0.0500 M HCl. Calculate the pH after addition of the following volumes of acid.

1.) 64 mL

2.) 89.4 mL

3.) 92 mL

Answer #1

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

When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which
of the following is correct for this titration?
A. Initially the pH will be less than 1.00.
B. The pH at the equivalence point will be 7.00.
C. It will require 12.50 mL of NaOH to reach the equivalence
point.
When 0.100 M NaOH is titrated with 25.00 mL 0.0500 M HBr, which
of the following is correct for this titration?
A. Initially the pH will be...

A 40.0 mL sample of 0.150 M Ba(OH)2 is titrated with 0.400 M
HNO3. Calculate the pH after the addition of the following volumes
of acid.
a) 0.0 mL
b) 15.0 mL
c) At the equivalence point
d) 40.0 mL

40.0 ml of 0.600M hydrofluoric acid, HF, is titrated with 0.400M
sodium hydroxide, NaOH. Calculate the pH of the solution after the
addition of 35.0 ml of NaOH solution.

a 30ml volume of .1M HCl is titrated with .05 NaOH. Calculate
the pH after the addition of the following volumes of base: 0ml,
40ml, 60ml, 80ml
Draw and label a titration curve

1. A 100.0-mL sample of 0.500 M sodium hydroxide is titrated
with 0.100 M nitric acid. Calculate the pH before the titration
begins.
2. A 1.0-L buffer solution contains 0.100 mol HCN and 0.100 mol
LiCN. The value of Ka for HCN is 4.9 x 10-10. Because the initial
amounts of acid and conjugate base are equal, the pH of the buffer
is equal to pKa = -log (4.9 x 10-10) = 9.31. Calculate the new pH
after the addition...

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....

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...

a) Calculate the volume of 0.0500 M NaOH needed to neutralize a
mixture of 25.00 mL of 0.05 M HCl and 20.00 mL of 0.05 M NH4Cl.
b) Calculate volume needed if 0.50 g of mannitol is added to
acid mixture.

Calculate the pH after 1.0 mL of 0.100 M HCl has been
added to 40.0 mL of 0.100 M piperazine.
(pKa1 = 5.333 and pKa2 =
9.731.)

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