Question

What is the pH at the equivalence point when 55.0 mL of a 0.300 M
solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to
its end point?

Answer #1

Since this isia weak acid- base titration so,

CH3COOH having 55.0ml(. 055 l) and .3M

for mole, . 055* . 3=0.0165 mole

So for NaOH 0.1650 l mixed with 0.1M soln to have 0.0165 mole of NaOH

NaOH + CH3CHOOH = NaCH3COO + H2O

Initial 0.0165 0.0165 0

Equilibrium 0 0 0.0165

Now Molarity of NaCH3COO = mole/litre

= 0.076

Now,

NaCH3COO + H2O = NaCH3COOH + OH-

Initial 0.075 0 0

Eqm 0.075-x x x

Kb = [NaCH3COOH] [OH-] /[NaCH3COO]

Kw=Kb*ka here ka for given acid is = 1.8*10^-10 and kw= 10^-14 so kb= 5.6*10^-10

Therefore, 5.6*10^-10 = x^2/0.075 since 0.075-x = 0.075

So x= 6.48*10^-6

Now pOH = - Log[6.48*10^-6]

= 5.188

Therefore pH = 14-5.188

= 8.812 answer

1. When 50.00 mL of 0.100 M NaOH reacts with 25.00 mL
of 0.300 M HX, a weak monoprotic acid, the pH of the resulting
solution is 3.74. Calculate the Ka of the acid.
2. Will the equivalence point of ammonia plus HCl be
higher, lower, or equal to 7? Explain.

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)

What is the pH at the EQUIVALENCE POINT of the titration of
0.100 M NaOH into a solution of 12.5 mL of 0.243 M butanoic
acid

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

A 300.0 mL buffer solution is 0.300 M in acetic acid and 0.300 M
in sodium acetate.What is the pH after addition of 0.0150 mol of
HCl? What is the pH after addition of 0.0150 mol of NaOH?

A
22.5 mL sample of an acetic acid solution is titrated with a 0.175M
NaOH solution. The equivalence point is reached when 37.5 mL of the
base is added.
What was the concentration of acetic acid in the original
(22.5mL) sample?
What is the pH of the equivalence point?
Ka acetic acid= 1.75E-5

What is the pH at the equivalence point when 50.00 mL of 0.112
M hydroxyacetic acid is titrated with 0.0580 M
KOH? (Assume Ka = 1.47???10?4, and
Kw = 1.01???10?14.)

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?

25.00 mL of 0.1 M acetic acid are titrated with 0.05 M NaOH.
Calculate the pH of the solution after addition of 25 mL of NaOH.
Ka(CH3COOH) = 1.8 x 10-5.

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