Question

Calculate the change in pH to 0.01 pH units caused by adding 10. mL of 3.69-M NaOH is added to 530. mL of each of the following solutions.

a.) water

pH before mixing: 7.00 (correct)

pH after mixing: ??

b.) 0.180 M NH₃

pH before mixing: 11.26

pH after mixing: ??

Thank you! Sincerely, a confused college student

Answer 1

a) Distilled water will have a pH of 7

pH when 10ml of 3.69M NaOH is added

NaOH----> Na+ +OH-

moles in 10ml of 3.69M NaOH= 3.69*10/1000 =0.0369 moles

when 530ml of water is mixed, volume =530+10 =540ml =0.54L

Concentration of NaOH = 0.0369/0.54 =0.0683M

NaOH is strong base and ionizes completely

[OH-] =0.0683M

p[OH-] =-log(0.0683)=1.17

pH= 14-1.17=12.83

2. NH3+H2O----> NH4+ +OH-

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

at equilibrium [NH3] =0.18-x and [NH4+] =[OH-] =x

x²/(0.18-x)= Kb= 1.8*10-5

looking at the magniture if Kb

0.18-x can be approximated to 0.18

x²= 1.8*0.18*10^-5 =

x=0.0018

pOH= -log(0.0018)= 2.74, pH= 14-2.74= 11.26

2. When NaOH is added, moles of NaOH= 0.00369 moles

Moles of NH3 in = 530*0.18/1000=0.0954

Total volume after mixing = 540 ml

NaOH concentration = 0.00369/0.54 =0.0683M

Concentratino of NH3= 0.0954/0.54=0.176 =0.18

So [OH-] due to NH₃ will not change much and is = 0.0018

NaOH ionzes completely to Na+ and OH- and [OH-] = 0.0683

total [OH-] =0.0018+0.0683=0.00863

p[OH-] =-log(0.00863)=2.06

pH= 14-2.06=11.94