Question

In: Chemistry

A 1 liter sample of a lake water has an Alk of .2 meq/liter and contains...

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 ?

Solutions

Expert Solution

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.


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