For 500.0 mL of a buffer solution that is 0.185 molL−1 in C2H5NH2 and 0.175 molL−1...

For 500.0 mL of a buffer solution that is 0.185 molL−1 in C2H5NH2 and 0.175 molL−1 in C2H5NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Expert Solution

Method One:

a)

. . . . . . . . . . . . . HONH2 + H2O ⇌ HONH3+ + OH-
Before add HCl: 0.400 . . . . . . . . . . 0. . . . .........trace
After add HCl: . .0.380 . . . . . . . . . . 0.020 . . . . .trace
Kb = [HONH3+]*[OH-]/[HONH2] = (0.020/0.380)*[OH-] =1.1E-8
[OH-] = 0.209E-6... ->...pOH = 6.68
pH+pOH = 14 ...->...pH = 7.32

b) HONH3Cl is the acid in this conjugated acid-base couple. Hence the added HCl will not react with HONH3Cl and [H3O+] = 0.020.

[H3O+] = 0.020. Therefore pH = -log(0.020) = 1.70

c)
mixture containing 0.400 M HONH2 and 0.400 M HONH3Cl
. . . . . . . . . . . . . HONH2 + H2O ⇌ HONH3+ + OH-
Before add HCl: 0.400 . . . . . . . . . . 0.400 . . . . .trace
After add HCl: . .0.380 . . . . . . . . . . 0.420 . . . . .trace
Kb = [HONH3+]*[OH-]/[HONH2] = 1.105E-8[OH-]=1.1E-8
[OH-] = 0.995E-8 -> pOH = 8.00
pOH +pH = 14 -> pH = 6.00

d) [H3O+] = 0.020. Therefore pH = -log(0.020) = 1.70

Method Two:

a)
pure water : pH = 7.00
moles NaOH = 0.010
[OH-]= 0.010 mol/ 0.240 L=0.0417 M
pOH = - log 0.0417= 1.38
pH = 14 - pOH = 14 - 1.38=12.6

b)
Ka = 1.8 x 10^-4
pKa = 3.74
pH = 3.74 + log 0.315 / 0.195 =3.95 ( initial pH of the buffer )

moles formic acid = 0.195 M x 0.240 L=0.0468
moles formate = 0.315 M x 0.240 L =0.0756
the effect of the added 0.010 mol OH- would be to decrease the moles of formic acid by 0.010 and increases the moles of formate by 0.010 by the reaction
HCOOH + OH- = HCOO-
moles HCOOH = 0.0468 - 0.010 = 0.0368
moles HCOO- = 0.0756 + 0.010=0.0856
concentration HCOOH = 0.0368 / 0.240 L=0.153 M
concentration HCOO- = 0.0856 / 0.240 =0.357
pH = 3.74 + log 0.357 / 0.153 = 4.11 ( final pH)

c)
Kb of ethylammine = 4.3 x 10^-4
pKb = 3.37
pOH = 3.37 + log 0.255 / 0.275 = 3.34
pH = 14 - 3.34 =10.7 ( initial pH)
moles CH3CH2NH3+ = 0.255 x 0.240 L=0.0612
moles CH3CH2NH2 = 0.275 x 0.240 L=0.0660
CH3CH2NH3+ + OH- = CH3CH2NH2 + H2O
moles CH3CH2NH3+ = 0.0612 - 0.010 =0.0512
moles CH3CH2NH2 = 0.0660 + 0.010 =0.0760
concentration CH3CH2NH3+ = 0.0512 / 0.240 =0.213 M
concentration CH3CH2NH2 = 0.0760 / 0.240=0.317 M
pOH = 3.37 + log 0.213 / 0.317=3.20
pH = 10.8

queen_honey_blossom answered 2 years ago

For 550.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.175 M...

For 550.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.175 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl. Express your answers using two decimal places separated by a comma.

A 120.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.135 molL−1 in NH4Br. If...

A 120.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.135 molL−1 in NH4Br. If the same volume of the buffer were 0.255 molL−1 in NH3 and 0.400 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00? the answer is not 0.084g!

A 110.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.125 molL−1 in NH4Br. (A)...

A 110.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.125 molL−1 in NH4Br. (A) What mass of HCl will this buffer neutralize before the pH falls below 9.00? (Kb(NH3)=1.8×10−5)? (B) If the same volume of the buffer were 0.265 molL−1 in NH3 and 0.390 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

A 130.0 mL buffer solution is 0.105 molL−1 in NH3 and 0.135 molL−1 in NH4Br. What...

A 130.0 mL buffer solution is 0.105 molL−1 in NH3 and 0.135 molL−1 in NH4Br. What mass of HCl will this buffer neutralize before the pH falls below 9.00? (Kb(NH3)=1.8×10−5). If the same volume of the buffer were 0.270 molL−1 in NH3 and 0.400 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00.

A 110.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.125 molL−1 in NH4Br. Part...

A 110.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.125 molL−1 in NH4Br. Part A (PLEASE ANSWER EVERY PART CORRECTLY :)) What mass of HCl will this buffer neutralize before the pH falls below 9.00? (Kb(NH3)=1.76×10−5 Express your answer using two significant figures. mm = g Part B If the same volume of the buffer were 0.265 molL−1 in NH3 and 0.390 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?...

#6. A 130.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.125 molL−1 in NH4Br....

#6. A 130.0 mL buffer solution is 0.110 molL−1 in NH3 and 0.125 molL−1 in NH4Br. A) What mass of HCl will this buffer neutralize before the pH falls below 9.00? (Kb(NH3)=1.76×10−5) B)If the same volume of the buffer were 0.270 molL−1 in NH3 and 0.390 molL−1 in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

PART 1: For 500.0 mL of a buffer solution that is 0.155 M in HC2H3O2 and...

PART 1: For 500.0 mL of a buffer solution that is 0.155 M in HC2H3O2 and 0.145 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.020 mol of HCl. ... PART 2: For 500.0 mL of a buffer solution that is 0.150 M in CH3CH2NH2 and 0.135 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

A 250.0-mL buffer solution is 0.290 molL−1 in acetic acid and 0.290 molL−1 in sodium acetate....

A 250.0-mL buffer solution is 0.290 molL−1 in acetic acid and 0.290 molL−1 in sodium acetate. Ka(CH3COOH)=1.8×10−5 Part A What is the initial pH of this solution? Express your answer using two decimal places. Part B What is the pH after addition of 0.0150 mol of HCl? Express your answer using two decimal places. Part C What is the pH after addition of 0.0150 mol of NaOH? Express your answer using two decimal places.

Acid Base Questions PART 1: For 500.0 mL of a buffer solution that is 0.150 M...

Acid Base Questions PART 1: For 500.0 mL of a buffer solution that is 0.150 M in CH3CH2NH2 and 0.135 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 molof HCl. PART 2: A buffer contains significant amounts of acetic acid and sodium acetate. Write an equation showing how this buffer neutralizes added base (KOH).

Part B) For 490.0 mL of a buffer solution that is 0.175 M in HC2H3O2 and...

Part B) For 490.0 mL of a buffer solution that is 0.175 M in HC2H3O2 and 0.155 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.020 mol of HCl. Part C) For 490.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.165 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Question