Question

Calculate the volume of 1M solution of HCl needed for 1 unit shift of the pH of the following solutions:

a) 100mL of 0.01M solution of weak acid (pKa=6) and 0.05M of it's sodium salt

b) 0.005M of sodium formate and 23g of formic acid dissolved in 100mL of water (pKa of formic acid=3.75)

c) 100mL of 0.1M ammonia solution

Answer 1

a)

moles of acid = molarity x volume / 1000

moles of acid = 0.01 x 100 /1000 = 10-3

moles of salt = 0.05 x 100/100 = 5 x 10-3

for acidic buffer

pH = pKa + log [ salt / acid ]

pH = 6 + log [ 0.05 / 0.01]

pH = 6.7

after adding HCl

Hcl is a strong acid so pH decreases

new pH = 6.7 -1 = 5.7

so

5.7 = 6 + log [ salt / acid ]

[salt] / [acid] = 0.5

as volume is same

ratio of conc = ratio of moles

moles of salt / moles of acid = 0.5

moles of salt = 0.5 moles of acid

sodium salt + Hcl ---> acid + NaCl

let the moles of HCl be y

so

new moles of acid = ( y + 10-3 )

new moles of salt = ( 5 x 10-3 - y )

so

( 5 x 10-3 - y ) = 0.5 ( y + 10-3 )

4.5 x 10-3 = 1.5 y

y = 3 x 10-3

moles of Hcl = 3 x 10-3

volume = moles / molarity

volume = 3 x 10-3 / 1 = 3 x 10-3 L = 3 ml

so 3 ml of 1 M HCl should be added

b)

moles of formic acid = 23 / 46 = 0.5

moles of sodium formate = 0.005

for acidic buffer

pH = pKa + log [ salt / acid ]

as volume is same

ratio of conc = ratio of moles

pH = 3.75 + log [ 0.005/ 0.5]

pH = 1.75

after adding HCl

Hcl is a strong acid so pH decreases

new pH = 1.75 -1 = 0.75

so

0.75 = 3.75 + log [ salt / acid ]

[salt] / [acid] = 10-3

as volume is same

ratio of conc = ratio of moles

moles of salt / moles of acid = 10-3

moles of salt = 10-3 moles of acid

sodium formate + Hcl ----. formic acid + NaCl

let the moles of HCl be y

so

new moles of acid = ( y + 0.5)

new moles of salt = ( 0.005 - y )

so

( 0.005 - y ) = 10-3( y + 0.5 )

5 - 1000y = y + 0.5

y = 4.5 x 10-3

moles of Hcl = 4.5 x 10-3

volume = moles / molarity

volume = 4.5 x 10-3 / 1= 4.5 x 10-3 L = 4.5 ml

so 4.5 ml of 1 M HCl should be added

c)
moles of NH3 = 0.1 x 100 / 1000 = 10 x 10-3

fow weak base

[OH-] = sqrt ( Kb C )

[OH-] = sqrt ( 0.1 x 1.778 x 10-5 )

[OH-] = 1.33 x 10-3

pOH = -log 1.33 x 10-3

pOH = 2.875

pH = 14 - 2.875 = 11.125

as Hcl decreases the pH

new pH = 10.125

new pOH = 3.875

Nh3 + HCl ---> Nh4Cl

let the moles of Hcl added be y

new NH3 moles = 10 x 10-3 - y

new Nh4Cl moles = y

it forms a basic buffer

pOH = pKb + log [ Nh4Cl / Nh3 ]

3.875 = 4.75 + log [ NH4Cl / NH3]

[NH4Cl] / [NH3] = 0.1333

y / ( 10 x 10-3 - y ) = 0.1333

y= 1.17667 x 10-3

so moles of Hcl is 1.17667 x 10-3

volume = volume / molarity

volume = 1.17667 x 10-3 / 1

volume = 1.17667 L = 1.177 ml

so the volume should of HCl should be 1.177 ml