Farmers who raise cotton once used arsenic acid, H3AsO4, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with K1 = 2.5 × 10–4, K2 = 5.6 × 10–8, and K3 = 3 × 10–13. What is the pH of a 0.500 M solution of arsenic acid?

pH = 1.95

Explanation:

For polyprotic acids, it is generally assumed that all H⁺ comes from the 1st ionization step. The amount of H⁺ delivered into solution for the 2nd and 3rd ionization steps are in the order of 10⁻⁴M and 10⁻⁶M respectively and provide very little change in pH from the quantity delivered in the 1st ionization step.

Therefore... the [H⁺] concentraion and pH are computed as follows...

[H⁺] = √Ka₁[H₃AsO₄] = √(2.5 x 10⁻⁴)(0.500) M = 0.1118M

pH = -log[H⁺] = -log(0.01118) = 1.95

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

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