50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, Ka=1.91.pdf

50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, Ka=1.9*10^-5) is titrated with a
0.1000M solution of sodium Hydroxide. Calculate the pH at the following points a. Before
Titration b. After the addition of 5.00mL of NaOH c. After the addition of 25.00mL of NaOH d.
After the addition of 50.00mL of NaOH e. After the addition of 5.00mL of NaOH
Solution
HN3+ NaOH--> NaN3 +H2O
mini moles of HN3=5
ka=Ka=1.9*10^-5
pka=4.72
a) ph before addition
before addition only weak acid is present
alpha=sqrt(ka/c)
=1.37*10^-2
c.alpha= [H+]= 1.37*10^-3
ph=2.86
b) after addition of 5 ml
mini moles of naoh added=0.5
ph=4.72+log(0.5/(5-0.5))
ph=3.76
c) after addition of 25 ml Naoh
ph=4.72+log(2.5/(5-2.5))
ph=4.72
d)after addition of 50 ml
now this is equivalence point.
Here N1v1=N2v2
here salt hydrolysis will take place
N3- +H2O HN3+OH-
pOH=1/2(pKw-pKa-logC)
C= 5/(50+50)
=1/20M
pOH=1/2(14-4.72+1.30)
=5.29
ph=8.71.

1. 50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, Ka=1.9*10^-5) is titrated with a
0.1000M solution of sodium Hydroxide. Calculate the pH at the following points a. Before
Titration b. After the addition of 5.00mL of NaOH c. After the addition of 25.00mL of NaOH d.
After the addition of 50.00mL of NaOH e. After the addition of 5.00mL of NaOH

Solution

HN3+ NaOH--> NaN3 +H2O
mini moles of HN3=5
ka=Ka=1.9*10^-5
pka=4.72

a) ph before addition
before addition only weak acid is present
alpha=sqrt(ka/c)
=1.37*10^-2
c.alpha= [H+]= 1.37*10^-3
ph=2.86
b) after addition of 5 ml
mini moles of naoh added=0.5
ph=4.72+log(0.5/(5-0.5))
ph=3.76
c) after addition of 25 ml Naoh
ph=4.72+log(2.5/(5-2.5))
ph=4.72
d)after addition of 50 ml
now this is equivalence point.
Here N1v1=N2v2
here salt hydrolysis will take place
N3- +H2O <=> HN3+OH-
pOH=1/2(pKw-pKa-logC)
C= 5/(50+50)
=1/20M
pOH=1/2(14-4.72+1.30)
=5.29

2. ph=8.71