Question

A solution was produced by dissolving NH3 to give a 250. mL solution with 3.00 M NH3. a. What is the pH of this solution? pH =

b. What mass of NH4Cl must be added to this solution to give a pH of 9.45? __ g

c. If 40. mL of 2.0 M KOH was added to this solution, what will be the pH of this solution? pH =

d. If 40. mL of 2.0 M KOH was added to the same volume (250. mL) of water, what will the pH be of this solution? pH=

Answer 1

Note that NH3 is a wek base, will hydrolyse to form

NH3 + H2O = NH4+ + OH-

a)

pH of V = 250 mL of 3 M of NH3,

Kb = [NH4+][OH-]÷[NH3]

Kb = 1.8*10^-5 for NH3

in equilibrium

[NH4+] = x

[OH-] = x

[NH3] = 3-x

substitute in Kb

1.8*10^-5 = x*x/(3-x)

x = OH- = 0.007339 M

pOH = -log(IOH) = -log(0.007339)

pH = 14-pOH = 14-2.1343

pH = 11.8657, which is basic as expected

b)

if we need pH = 9.45

use henderson hasselback equation, since there will be:

weak base + conjugate acid; or NH3 and NH4+ respectively

from Hendesron hasselbach equation

pH = pKa + log(NH3/NH4+)

pKa can be calcualted via Kb:

pKa = 14-pKb = 14-log(Kb) = 14-log(1.8*10^-5) = 9.25

pH = 9.45 and NH3 is given as 3 M, NH4+ must be calcualtedd

substitute all this data

9.45 = 9.25 + log(3/([NH4+]))

10^(9.45-9.25) = 3/([NH4])

[NH4+] = 3 /( 1.584 ) = 1.89393 M

now, get NH4Cl

[NH4Cl] = [NH4+]= 1.89393 M (stoichiometry)

mol = M*V = (1.89393)(0.25) = 0.47348 moles of NH4Cl

mass = mol*MW= (0.47348)(53.491) = 25.326 g of NH4Cl

c)

assume we are adding this to the buffer solution (the one in b)

so

mmol of KOH = MV = 40*2 = 80 mmol of OH-

this will react with the conjugate acid, NH4+

mmol of NH4+ = 0.47348 mol = 0.47348 = 473.48 mmol of NH4+

mmol of NH3 = 3*250 mL = 750 mmol

after reaction of OH-

mmol of NH4+ =473.48-80 = 393.48

mmol of NH3 = 750+ 80 = 830

substitute in pH equation

pH = 9.25 + log(NH3/NH4+)

pH = 9.25 + log(830 / 393.48 ) =9.5741

d)

if we only added this to V = 250 mL of water:

V total = 250+40 = 290 mL

mmol of KOH = MV = 80 mmol from previous data

[KOH] = mmol / V = 80/290 = 0.2758 M

pOH = -log(0.2758) = 0.559405

pH = 14-pOH = 14-0.559405 = 13.443