40.0-ml Sample Of 0.0100 M Hc2h3o2 Is Titrated With 0.0200 M Naoh. Calculate The Ph After The

A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Determine the pH at the equivalence point of ti?

at the equivalence point it's established a weak base, NO2^-

pH ca equal 10,5 to 11 at eq point

Calculate the pH after the addition of 4.00mL of 0.100M HCl?

For each of the following three questions 25.0 mL of 0.100 M pyridine (Kb = 1.70 x 10−9) is titrated with 0.100 M HCl. For each question calculate the pH of the final solution after the indicated amount of HCl has been added. That is, treat each question as a separate titration.

Pyridine is monobasic, i.e., each molecule of pyridine accepts only one proton from a molecule of acid. Remember that the final volume of the solution is the sum of the initial volume and the volume of HCl that is added to it.

Calculate the pH after the addition of 4.00 mL of 0.100 M HCl.
Calculate the pH after the addition of 25.0 mL of 0.100 M HCl.
Calculate the pH after the addition of 28.0 mL of 0.100 M HCl.

A 25.0 mL sample of 0.100 M acetic acid is titrated with 0.125 M NaOH Calculate the pH of titration mixture?

.095672897636479949575584832848382483248...

Calculating pH during titration?

100.0 mL of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide. An appropriate indicator is used. Ka for acetic acid is 1.7 s 10^-5. Calculate the pH in the flask at the following points in the titration.

a) When no NaOH has been added
b) After 25.0 ml of NaOH is added
c) After 50.0 ml of NaOH is added
d) After 75.0 ml of NaOH is added
e) After 100.0 ml of NaOH is added
f) After 300.0 ml of NaOH is added

I just want to make sure that my math is right...

For A- I got a pH of 1 by doing the -log of ((.100 M x .100 L) - (.100 M x 0 L)) // .100 L.

I used the same technique to get
b) 1.22
c) 1.48
d) 1.85

Is that correct? And then I'm having a hard time with e because with what I'm doing I get -log(0) which of course is undefined....

Help would be appreciated!

A 20.0-mL sample of 0.25 M HNO3 is titrated with 0.15 M Na OH. What is the pH of the solution after 30.0 mL of

Moles HNO3 = 0.020 L x 0.25 =0.0050
Moles NaOH = 0.030 L x 0.15 = 0.0045

Moles HNO3 in excess = 0.0050 - 0.0045 =0.00050

total volume = 0.050 L

concentration H+ = 0.00050 / 0.050 = 0.0100 M

pH = 2.00

Calculating pH at the Equivalence Point?

Moles HBr = 0.096 L x 0.25 =0.024
Moles KOH = 0.0480 L x 0.50 =0.024
Moles strong acid = moles strong base => pH = 7

Moles ammonia = 0.0500 x 0.20 =0.0100
Moles HNO3 = 0.0500 x 0.20 = 0.100
NH3 + H+ >> NH4+
total volume = 0.100 L
[NH4+] = 0.0100 mol / 0.100 L =0.100 M
NH4+ + H2O <----> NH3 + H3O+
K = Kw/ Kb = 5.6 x 10^-10 = x^2 / 0.100-x
x = [H+] =7.5 x 10^-6 M
pH = 5.1

Moles acid = 0.030 L x 0.50 = 0.015
moles NaOH = 0.030 x 0.50 = 0.015
CH3COOH + OH- >> CH3COO- + H2O
total volume = 0.0600 L
[CH3COO-] = 0.015 / 0.0600 =0.25 M
CH3COO- + H2O <---> CH3COOH + OH-
K = Kw/ Ka = 5.6 x 10^-10 = x^2 / 0.25-x
x = [OH-] = 1.2 x 10^-5 M
pOH = 4.9
pH = 9.1

25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH at the equivalence point?

OH- (aq) + CH3COOH (aq) ---> CH3COO- (aq) + H2O (l)

At equivalence, the solution is an aqueous salt. So, the pH is determined from the equilibrium of a weak base, the conjugate base of HA
A-(aq) + H2O = OH-(aq) + HA(aq)

The equilibrium constant of this reaction is related to the pKa of the acid HA and the solution of the equilibrium expression for this reaction will yield the hydroxide concentration, [OH-(aq)].


We may use the concentrated weak base approximation, in exact analogy to the concentrated weak acid approximation, to determine the [OH-(aq)], whcih is tantemount to ignoring y in the denominator of the equilibrium expression. Therefore:

y = [OH-(aq)] = (Kb [A-])^1/2 = =sqrt(5.5*10^-10 * 0.05) = 5.25 x 10-6 M
And, of course, the pOH is then:

pOH = -log10([OH-(aq)]) = 5.3
and the pH is:

pH = 14 - pOH = 8.7