Question

Calculate the pH of the solution when 75mL of 0.092M has been titrated with 0, 15, 23, 30mL of 0.1M of NaOH.

Calculate the pH of the solution when 20mL of 0.1M CH3COOH with 0, 10, 20, 30mL of 0.1M of NaOH.

Answer 1

Solution:

pH of the solution when 75mL of 0.092M has been titrated with 0, 15, 23, 30mL of 0.1M of NaOH, in this molecule name not given and considered as acetic acid the computation are performed below,

for 0 ml,

millimoles of acetic acid =75 ml*0.092 M=6.9

millimoles of NaOH = 0*0.1M=0

mmoles left = 6.9-0=6.9

M of acetic acid = mmols/total volume = 6.9/75=0.092

pH=-log(0.092)=1.0362

for 15 ml,

millimoles of acetic acid =75 ml*0.092 M=6.9

millimoles of NaOH = 15*0.1M=1.5

mmoles left = 6.9-1.5=5.4

M of acetic acid = mmols/total volume=5.4/900.06=

pH=-log(0.06) = 1.22

for 23 ml,

millimoles of acetic acid =75 ml*0.092 M=6.9

millimoles of NaOH = 23*0.1M=2.3

mmoles left = 6.9-2.3=4.6

M of acetic acid = mmols/total volume=4.6/98=0.0469

pH=-log(0.0469)=1.328

for 30 ml,

millimoles of acetic acid =75 ml*0.092 M=6.9

millimoles of NaOH = 30*0.1M=3

mmoles left = 6.9-3=3.9

M of acetic acid = mmols/total volume=3.9/105=0.0371

pH=-log(0.0371)=1.43

the pH of the solution when 20mL of 0.1M CH3COOH with 0, 10, 20, 30mL of 0.1M of NaOH and its ccomputation is shown below,

for 0 ml,

millimoles of acetic acid =20*0.1M=2

millimoles of NaOH = 0*0.1M=0

mmoles left =2-0=2

M of acetic acid = mmols/total volume=2/20=0.10.1)=1

pH=-log(0.1)=1

for 10 ml,

for 0 ml,

millimoles of acetic acid =20*0.1M=2

millimoles of NaOH = 10*0.1M=1

mmoles left =2-1=1

M of acetic acid = mmols/total volume=1/30=0.033

pH=-log(0.0333)=1.477

for 20 ml,

for 0 ml,

millimoles of acetic acid =20*0.1M=2

millimoles of NaOH = 20*0.1M=2

mmoles left =2-2=0

M of acetic acid = mmols/total volume=0

pH=nil

for 30 ml,

for 0 ml,

millimoles of acetic acid =20*0.1M=2

millimoles of NaOH = 30*0.1M=3

mmoles left =2-3=-1

M of acetic acid = mmols/total volume=1/50=0.02

pH=-log(0.02)=1.69