Question

For the next 5 questions,

(a) please consider the titration of 20.00 mL of 0.405 M NH3 (a weak base) using 0.175 M HCl. What is the volume of HCl needed to react the equivalence point? Please answer only numerically in units of mL.

(b) For the titration of 20.00 mL of 0.405 M NH3 using 0.175 M HCl, what is the initial pH?

(c) For the titration of 20.00 mL of 0.405 M NH3 using 0.175 M HCl, what is the pH after 30.00 mL of HCl has been added?

(d) For the titration of 20.00 mL of 0.405 M NH3 using 0.175 M HCl, what is the pH after 46.28 mL of HCl has been added?

I need help with those questions, i want to make sure if i did them right? please help me as soon as possible.

Answer 1

a)

find the volume of HCl used to reach equivalence point

M(NH3)*V(NH3) =M(HCl)*V(HCl)

0.405 M *20.0 mL = 0.175M *V(HCl)

V(HCl) = 46.3 mL

Answer: 46.3 mL

b)when 0.0 mL of HCl is added

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.405 0 0

0.405-x x x

Kb = [NH4+][OH-]/[NH3]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.8*10^-5)*0.405) = 2.7*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.7*10^-3 M

So, [OH-] = x = 2.7*10^-3 M

use:

pOH = -log [OH-]

= -log (2.7*10^-3)

= 2.5686

use:

PH = 14 - pOH

= 14 - 2.5686

= 11.43

Answer: 11.43

c)when 30.0 mL of HCl is added

Given:

M(HCl) = 0.175 M

V(HCl) = 30 mL

M(NH3) = 0.405 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.175 M * 30 mL = 5.25 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.405 M * 20 mL = 8.1 mmol

We have:

mol(HCl) = 5.25 mmol

mol(NH3) = 8.1 mmol

5.25 mmol of both will react

excess NH3 remaining = 2.85 mmol

Volume of Solution = 30 + 20 = 50 mL

[NH3] = 2.85 mmol/50 mL = 0.057 M

[NH4+] = 5.25 mmol/50 mL = 0.105 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {0.105/5.7*10^-2}

= 5.01

use:

PH = 14 - pOH

= 14 - 5.01

= 8.99

Answer: 8.99

d)when 46.28 mL of HCl is added

Given:

M(HCl) = 0.175 M

V(HCl) = 46.28 mL

M(NH3) = 0.405 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.175 M * 46.28 mL = 8.099 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.405 M * 20 mL = 8.1 mmol

We have:

mol(HCl) = 8.099 mmol

mol(NH3) = 8.1 mmol

8.099 mmol of both will react

excess NH3 remaining = 0.001 mmol

Volume of Solution = 46.28 + 20 = 66.28 mL

[NH3] = 0.001 mmol/66.28 mL = 0 M

[NH4+] = 8.099 mmol/66.28 mL = 0.1222 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {0.1222/1.509*10^-5}

= 8.653

use:

PH = 14 - pOH

= 14 - 8.6532

= 5.3468

Answer: 5.35