In: Chemistry
A 1 liter sample of a lake water has an Alk of .2 meq/liter and contains .18 mM of carbonate species.
a. what is the pH ?
b. how much does pH change when 5 x 10 to the -5 power mol of hydrochloric acid are added ?
a) Base= 2 meq/lit = 210-3 eq/lit
CO32- = 0.18 mM =0.1810-3 eq
=( 0.182)10-3
= 0.910-3eq
Total Normality =(V1N1+V2N2)V
=[ (210-3)+(0.910-3)] 1000
= 2.910-6
[OH-] = 2.910-6
POH= -log[OH-]
= -log[2.910-6]
= 6 - log2.9
= 6 - 0.4623 = 5.537
PH = 14 - POH
= 14 - 5.537 =8.463
b) 510-3HCl added i.e 0.0510-3eq
Total normality =[ (2.9 - 0.05)10-3]103
= 2.8510-6
POH = - log [OH-]
= - log[2.8510-6]
= 6 - log 2.85 = 6 - 0.4548 = 5.5452
PH = 14 - 5.5452 = 8.4548 8.4545
Comare a & b PH VALUES 0.008 PH Change obtained.