In: Chemistry
50mL of a 0.1000 M aqueous solution of hydrazoic acid (HN3, pKa=4.72) 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 51.00mL of NaOH. PLEASE EXPLAIN STEPS AND NOT JUST SHOW WORK, I KNOW THE FIRST ONE, THE OTHERS ARE GIVING ME TROUBLE
a. Before Titration
pH = 1/2(pka-logC)
= 1/2(4.72-log0.1)
= 2.86
b. After the addition of 5.00mL of NaOH
no of mole of HN3 = 50*0.1/1000 = 0.005 mole
no of mole of NaOH = 5*0.1/1000 = 0.0005 mole
pka of HN3 = 4.72
pH = pka + log(base/acid)
= 4.72 +log(0.0005/(0.005-0.0005))
= 3.76
c. After the addition of 25.00mL of NaOH
no of mole of HN3 = 50*0.1/1000 = 0.005 mole
no of mole of NaOH = 25*0.1/1000 = 0.0025 mole
pka of HN3 = 4.72
pH = pka + log(base/acid)
= 4.72 +log(0.0025/(0.005-0.0025))
= 4.72
d. After the addition of 50.00mL of NaOH
no of mole of HN3 = 50*0.1/1000 = 0.005
mole
no of mole of NaOH = 50*0.1/1000 = 0.005 mole
concentration of salt formed = 0.005/100*1000 = 0.05 M
reaches to equivalencepoint
pH = 7+1/2(pka+logC)
= 7+1/2(4.72+log0.05)
= 8.71
e. After the addition of 51.00mL of NaOH
no of mole of NaOH excess = (51-50)*0.1/1000 = 0.0001 mole
concentration of NaOH excess = 0.0001/101*1000 = 0.00099 M
pH = 14 - (-log(OH-))
= 14 - (-log(0.00099))
= 11