In: Chemistry
Calculation of pH after Titration of Weak Acid: Calculation of pH after Titration of Weak Acid: A compound has a p K a of 7.4. To 100 mL of a 1.0 M solution of this compound at pH 8.0 is added 30 mL of 1.0 M hydrochloric acid. What is the pH of the resulting solution?
Can you explain what to do after 3.98 = A/HA
moles of HCl in 30ml of 0.1 = molarity* Volume (L)= 0.1*30/1000 =0.003 moles of acid in 100ml of 1M= 1*100/1000=0.1M
given pH= pKa+ log[A-]/[HA] where A- is conjugate base and HA is acid
8= 7.4 + log[A-]/[HA]
log [A-]/[HA] =0.6
[A-]/[HA] =3.98 , A- =3.98[HA]
A+ [HA] =1
4.98HA= 1, [HA] =1/4.98= 0.200
[A-] = 3.98*0.2 = 0.796 M
now : moles : [A-] =0.796*0.1= 0.0796, [HA]= 0.2*0.1=0.02 HCl =0.003
The addition of HCl (which ionizes completely) leads to formation of more HA.
moles of HA formed = 0.003+0.0796=0.799, moles of A- remaining = 0.0796-0.003= 0.793
now pH= 7.4+ log [0.793/130/(0.799/130) = 7.4+ log (0793/0.799)=7.396