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

CO₃^2- = 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

P^OH= -log[OH^-]

= -log[2.910^-6]

= 6 - log2.9

= 6 - 0.4623 = 5.537

P^H = 14 - P^OH

= 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

P^OH = - log [OH^-]

= - log[2.8510^-6]

= 6 - log 2.85 = 6 - 0.4548 = 5.5452

P^H = 14 - 5.5452 = 8.4548 8.4545

Comare a & b P^H VALUES 0.008 P^H Change obtained.

queen_honey_blossom answered 6 months ago

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