Question

In: Chemistry

1)A 21.7 mL sample of 0.267 M dimethylamine, (CH3)2NH, is titrated with 0.241 M perchloric acid....

1)A 21.7 mL sample of 0.267 M dimethylamine, (CH3)2NH, is titrated with 0.241 M perchloric acid. The pH before the addition of any perchloric acid is

2)A 27.1 mL sample of 0.212 M dimethylamine, (CH3)2NH, is titrated with 0.201 M hydroiodic acid. After adding 40.6 mL of hydroiodic acid, the pH is

Solutions

Expert Solution

dimethyl amine has pkb of 3.29 , Kb = 10 ^- 3.29 = 0.000513

(CH3)2NH moles = M x V ( in L) = 0.267 x 0.0217 = 0.005794

The base dissociation eq is given by

               (CH3)2NH (aq) + H2O (l)   <----> (CH3)2NH2+ (aq) +    OH- (aq)

Initial            0.005794                                            0                      0

equilibrium    0.005794-X                                         X                        X

Kb = [(CH3)2NH2+] [OH-] / [ (CH3)2NH]

0.000513 = ( X/0.0217)^2 / ( 0.005794-X) / (0.0217)

X^2 / ( 0.005794-X) = 0.0000111

X^2 + 0.0000111X - 6.45 x 10^-8 = 0

X = 0.0002485 , [OH-] = 0.0002485 /0.0217 = 0.01145

pOH = -log [OH-] = -log ( 0.01145) = 1.94

pH = 14-pOH = 14-1.94 = 12.06

2) HI acid moles added = M x V = 0.201 x 0.0406 = 0.00816

(CH3)2NH moles = 0.212 x 0.0217 = 0.0046

we had excess of acid compared to base

Acid and base react in 1:1 ratio

excess acid moles = 0.00816 -0.0046 = 0.00356

total solution volume = 27.1 + 40.6 = 67.7 ml = 0.0677 L

[HI] = moles / volume = 0.00356 / 0.0677 = 0.0526

[H+] = 0.0526       ( since 1HI gives 1H+ )

pH = -log [H+] = -log (0.0526) = 1.28


Related Solutions

A 24.3 mL sample of 0.349 M dimethylamine, (CH3)2NH, is titrated with 0.275 M hydrochloric acid....
A 24.3 mL sample of 0.349 M dimethylamine, (CH3)2NH, is titrated with 0.275 M hydrochloric acid. (1) Before the addition of any hydrochloric acid, the pH is _____ (2) After adding 13.6 mL of hydrochloric acid, the pH is _____ (3) At the titration midpoint, the pH is _____ (4) At the equivalence point, the pH is _____ (5) After adding 45.9 mL of hydrochloric acid, the pH is _____ Equilibrium constants are found here: https://docs.google.com/document/d/1cNVOoEOVMqPSx8QdL8mc3zd8yFTwBx8VWd2MiDCDA1o/edit
Determine the pH during the titration of 34.6 mL of 0.350 M dimethylamine ((CH3)2NH , Kb...
Determine the pH during the titration of 34.6 mL of 0.350 M dimethylamine ((CH3)2NH , Kb = 5.9×10-4) by 0.350M HCl at the following points. (a) Before the addition of any HCl-? (b) After the addition of 14.0 mL of HCl-? (c) At the titration midpoint -? (d) At the equivalence point -? (e) After adding 50.5 mL of HCl-?
A 29.6 mL sample of 0.260 M diethylamine, (C2H5)2NH, is titrated with 0.338 M hydrochloric acid....
A 29.6 mL sample of 0.260 M diethylamine, (C2H5)2NH, is titrated with 0.338 M hydrochloric acid. After adding 33.7 mL of hydrochloric acid, the pH is . Use the Tables link in the References for any equilibrium constants that are required.
A 25.6 mL sample of 0.219 M trimethylamine, (CH3)3N, is titrated with 0.373 M hydrochloric acid....
A 25.6 mL sample of 0.219 M trimethylamine, (CH3)3N, is titrated with 0.373 M hydrochloric acid. After adding 5.62 mL of hydrochloric acid, the pH is .
Calculate the pH of a 0.0333 M aqueous solution of dimethylamine ((CH3)2NH, Kb = 5.9×10-4) and...
Calculate the pH of a 0.0333 M aqueous solution of dimethylamine ((CH3)2NH, Kb = 5.9×10-4) and the equilibrium concentrations of the weak base and its conjugate acid. pH = [(CH3)2NH]equilibrium = M [(CH3)2NH2+ ]equilibrium = M
Calculate the pH of a 0.0438 M aqueous solution of dimethylamine ((CH3)2NH, Kb = 5.9×10-4) and...
Calculate the pH of a 0.0438 M aqueous solution of dimethylamine ((CH3)2NH, Kb = 5.9×10-4) and the equilibrium concentrations of the weak base and its conjugate acid. Calculate the pH of a 0.289 M aqueous solution of quinoline (C9H7N, Kb = 6.3×10-10) and the equilibrium concentrations of the weak base and its conjugate acid.
1). A 27.1 mL sample of 0.342 M trimethylamine, (CH3)3N, is titrated with 0.336 M hydrochloric...
1). A 27.1 mL sample of 0.342 M trimethylamine, (CH3)3N, is titrated with 0.336 M hydrochloric acid. The pH before the addition of any hydrochloric acid is _____ 2). A 20.8 mL sample of 0.305 M dimethylamine, (CH3)2NH, is titrated with 0.255 M hydroiodic acid. At the equivalence point, the pH is ______ 3). A 26.2 mL sample of 0.251 M diethylamine, (C2H5)2NH, is titrated with 0.215 M hydrobromic acid. After adding 42.8 mL of hydrobromic acid, the pH is...
Consider a 0.64 M solution of dimethylamine, (CH3)2NH (Kb = 5.4×10-4). Mark the major species in...
Consider a 0.64 M solution of dimethylamine, (CH3)2NH (Kb = 5.4×10-4). Mark the major species in the solution. _______TrueFalse (CH3)2NH _______TrueFalse (CH3)2NH2+ _______TrueFalse H2O _______TrueFalse H+ _______TrueFalse OH- Complete the following table (in terms of "x", the amount of dimethylamine which reacts). Minus signs must be included, omit positive signs and omit molarity units (they are assumed). (CH3)2NH (CH3)2NH2+ OH- Initial Change Equilibrium 0.64 - x Determine the equilibrium concentration of (CH3)2NH2+ M Calculate the pH of the solution. pH...
In the titration of 30.0 mL of 0.200 M (CH3)2NH with 0.300 M HI. Determine the...
In the titration of 30.0 mL of 0.200 M (CH3)2NH with 0.300 M HI. Determine the pH at the 1/4 of the waypoint in this titration. Determine the pH at the eqivalence point in this titration. Compare both answers and explain if they make sense. (Thanks for the help!) Sorry! The value for Ka is 3.2 x 10^9 and Kb is 5.4 x 10^-4
Dimethylamine, (CH3)2NH, is an organic base with pKb= 3.23at 298 K. In an experiment, a 40.0...
Dimethylamine, (CH3)2NH, is an organic base with pKb= 3.23at 298 K. In an experiment, a 40.0 mL sample of 0.105mol/L-1(CH3)2NH(aq)is titrated with 0.150mol/L-11HI(aq)at 298 K. (a) Write a balanced chemical equation for the neutralization reaction upon which this titration is based, and indicate clearly which atom is being protonated. Then, calculate the equilibrium constant for the neutralization reaction. (Hint: To calculate the equilibrium constant, you may find it helpful to represent the neutralization reaction as the sum of twoor morereactions.)...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT