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1.How many grams of benzoic acid (C_{6}H_{5}COOH) must be dissolved in 250 mL of water to have a solution with pH of 3.02. Ka=3x10^-5
your experiment, you need 2.1 L of a solution with a pH of 3.50. How many mL of 2 mM H_{2}SO_{4} solution you need to use to prepare the desired solution?
3. Calculate the pH, [H_{3}O+] and [S04] of 0.55 M H_{2}SO_{4} solution? (Kaz: 1.1x10-2)

1. Formula of weak acids pH = ½( pKa - log [c HA ] 0 )
2. Calculate cH+ of 2.1 l pH 3.5, this amount has to be delivered from 2 mM H_{2}SO_{4}.
3.We consider the example of sulfuric acid, cS = 0.01 mol/l.

1st stage

H_{2}SO_{4} => HSO_{4}- + H+

practically 100% - strong acid

2nd stage

HSO_{4}- => SO_{42}- + H+

Equilibrium, pKa = 1.92 - weak acid

The calculation of the equilibria (all particle types) leads to the following solution:
pH = 1.838; c(H_{2}SO_{4}) = 7.9*10-8 (0%); c(HSO_{4}-) = 5.5*10-3 (55%); c(SO_{42}-) = 4.5*10-3 (45%).
> A calculation would be wrong: cS = 0.01; pH = - log( cS ) = 2. (Only the first stage considered)
> One calculation is also wrong: Sulfuric acid is bivalent, so 1 H_{2}SO_{4} produces 2 H+;
for cS = 0.01 we have c( H+ ) = 2 * cS = 0.02; pH = - log( 0.02 ) = 1.699.
> A correct solution with "chemical considerations" would be:
The 1st stage of dissociation is complete; 0.01 mol/l H_{2}SO_{4} therefore produces 0.01 mol/l H+ and 0.01 mol/l HSO_{4}-.
For the 2nd stage we have an equilibrium: KS = 10-1.92 = 0.01202 = c( H+ ) * c( SO_{42}- ) / c( HSO_{4}- ).
If a portion x of HSO_{4}- dissociates: x SO_{42}- are formed, c( HSO_{4}- ) = 0.01 decreases by x
and c( H+ ) = 0.01 increases by x: KS = [ 0.01 + x ] * x / [ 0.01 - x ]
The (positive) solution of the quadratic equation x2 + ( 0.01 + KS ) * x - KS * 0.01 = 0 is: x = 0.004528.
From this follows the result: c( H+ ) = 0.014528; pH = 1.838; c(HSO_{4}-) = 5.5*10-3; c( SO_{42}- ) = 4.5*10-3.