Question

Consider the titration of 50.0 ml of 0.100 M methylamine (Kb = 5.0 × 10- 4) with 0.200 M HCl. What is the pH of the solution when 25.00 ml of HCL has been added?

Answer #1

In this titration we are using 50 ml of methylamine
(CH_{3}NH_{2}) with 0.200 M HCl

Moles of H^{+} = 0.025 L of HCl X 0.200 mol/L= 0.005
moles of H^{+}

Moles of methylamine= 0.05 L of methylamine X 0.100 M = 0.005 moles of methylamine

pH of solution

CH_{3}NH_{2} + H^{+}
----------------> CH_{3}NH^{+}_{3}

Initial 0.005 0.005 0

Final 0 0.005 0.005

Ignoring any hydronium ion are producing in the solution by the ionization of methyamine

H^{+} = 0.005 / 0.075 (total amount of solution) = 0.067
M

pH = - log 0.067 = 1.17

.Consider the titration of 100.0 mL of 0.100 M H2NNH2
(Kb=3.0E-6) by 0.200 M HNO3. Calculate the pH of the resulting
solution after the following volumes of HNO3 have been added.
A) 0.0 mL B) 20.0 mL C) 25.0 mL D) 40.0 mL E) 50.0 mL F)100.0
mL

Consider the titration of 100.0 mL of 0.200 M CH3NH2 by 0.100 M
HCl. For each volume of HCl added, decide which of the components
is a major species after the HCl has reacted completely. Kb for
CH3NH2 = 4.4 x 10-4. 1) Calculate the pH at the equivalence point
for this titration 2) At what volume of HCl added in this titration
does the pH = 10.64? Express your answer in mL and include the
units in your answer.

Consider titration of 100.0 mL of 0.100 M H2NNH2 (Kb =
3.5x10^-6) by 0.200 M HNO3? Calculate the pH of the resulting
solution after the following volumes of HNO3 have been added.
0.0 mL
20.0 mL
25.0 mL
40.0 mL
50.0 mL
100.0 ml
Can anyone answer this thouroughly please? Please be sure to
show every step of the math, that us were I tend to get lost. Thank
you!

Calculate the pH at the equivalence point in the titration of
50.0 mL of 0.125 M methylamine (Kb = 4.4 × 10−4) with 0.265 M
HCl..

During the titration of 25.00 mL of 0.130 M weak base NH3 (Kb =
1.8 x 10^-5) with 0.100 M HCl, calculate the pH of the solution at
the following volumes of acid added:
a) 10.00 mL
b) 32.50 mL
c) 50.00 mL

Calculate the pH during the titration of 20.00 mL of 0.1000 M
methylamine, (CH3)NH2(aq), with 0.2000 M HCl(aq) after 5.5 mL of
the acid have been added. Kb of methylamine = 3.6 x 10-4. Answer =
10.47
Please show process.

In the titration of a 25.00 mL of 0.245 M weak base (Kb= 1.0
*10^-4) being titrated by 0.365 M HCl determine the pH at
equivalence point and the pH after 22.4 mL of HCl has been reached.
Please explain! I'm having a hard time with these type of
questions.

Calculate the pH at the equivalence point for the titration of
0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of
methylamine is 5.0× 10–4.
pH=?

Calculate the pH at the equivalence point for the titration of
0.130 M methylamine (CH3NH2) with 0.130 M HCl. The Kb of
methylamine is 5.0× 10–4.

Calculate the pH at the equivalence point for the titration of
0.240 M methylamine (CH3NH2) with 0.240 M HCl. The Kb of
methylamine is 5.0× 10–4.

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