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Question: \({{H}_{3}}P{{O}_{4}}\) is tribasic acid with \(p{{K}_{a1}},p{{K}_{a2}}\text{ }and\text{ }p{{K}_{a3}...

H3PO4{{H}_{3}}P{{O}_{4}} is tribasic acid with pKa1,pKa2 and pKa3p{{K}_{a1}},p{{K}_{a2}}\text{ }and\text{ }p{{K}_{a3}} 2.12, 7.21 and 12.32 respectively. It is used in fertilizer productions and its various salts are used in food, detergent, toothpaste, and in mineral treatment.
Small quantities of H3PO4{{H}_{3}}P{{O}_{4}} are used in imparting the sour or tart taste to soft drinks, such as coca cola, and beers, in which H3PO4{{H}_{3}}P{{O}_{4}} is present 0.05 % by weight (density = 1.0 g/ml). 10-3 M H3PO4{{H}_{3}}P{{O}_{4}} (pH=7) is used in fertilizers as an aqueous soil digesting. Plants can absorb zinc in water soluble form only. Zinc phosphate is the source of zinc and PO32−PO_{3}^{2-} ions in the soil,  Ksp~{{K}_{sp}} of zinc phosphate = 9.1 ×10−339.1\text{ }\times {{10}^{-33}} , [Zn2+]\left[ Z{{n}^{2+}} \right] ion in the soil is:
A. 2.9×10−11M2.9\times {{10}^{-11}}M
B. 4.0×10−10M4.0\times {{10}^{-10}}M
C. 3.0×10−6M3.0\times {{10}^{-6}}M
D. 9.1×10−5M9.1\times {{10}^{-5}}M

Explanation

Solution

. First we have to calculate the Ka1,Ka2 and Ka3{{K}_{a1}},{{K}_{a2}}\text{ }and\text{ }{{K}_{a3}} from pKa1,pKa2 and pKa3p{{K}_{a1}},p{{K}_{a2}}\text{ }and\text{ }p{{K}_{a3}} by using the following relationship between Ka{{K}_{a}} and pKap{{K}_{a}} .
Ka=10−pKa{{K}_{a}}={{10}^{-p{{K}_{a}}}}
Here, Ka{{K}_{a}} = acid dissociation constant.
pKap{{K}_{a}} = ionization constant.

Complete step by step answer:
- In the question it is given that pH of the solution is 7.
- Then [H3O+]=10−7[{{H}_{3}}{{O}^{+}}]={{10}^{-7}}
- In the question it is given that the pKa1,pKa2 and pKa3p{{K}_{a1}},p{{K}_{a2}}\text{ }and\text{ }p{{K}_{a3}} values are 2.12, 7.21 and 12.32 respectively.
- The Ka1,Ka2 and Ka3{{K}_{a1}},{{K}_{a2}}\text{ }and\text{ }{{K}_{a3}} values will be 7.59×10−3,6.17×10−8and4.79×10−137.59\times {{10}^{-3}},6.17\times {{10}^{-8}}and4.79\times {{10}^{-13}} respectively.

- Therefore for tribasic acid
=Ka1 Ka2 Ka3Ka1 Ka2 Ka3+Ka1 Ka2 !![!! H3O+ !!]!! +Ka1 !![!! H3O+ !!]!! 2+ !![!! H3O+ !!]!! 3=\dfrac{{{K}_{a1}}\text{ }{{K}_{a2}}\text{ }{{K}_{a3}}}{{{K}_{a1}}\text{ }{{K}_{a2}}\text{ }{{K}_{a3}}+{{K}_{a1}}\text{ }{{K}_{a2}}\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ +}{{K}_{a1}}\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}{{\text{ }\\!\\!]\\!\\!\text{ }}^{\text{2}}}+{{\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }}^{3}}}

- Substitute all the known values in the above equation to get the value.
f=Ka1 Ka2 Ka3Ka1 Ka2 Ka3+Ka1 Ka2 !![!! H3O+ !!]!! +Ka1 !![!! H3O+ !!]!! 2+ !![!! H3O+ !!]!! 3 =7.59×10−3×6.17×10−8×4.79×10−137.59×10−3×6.17×10−8×4.79×10−13+7.59×10−3×6.17×10−8×10−7+7.59×10−3×6.17×10−8×10−7+7.59×10−3×(10−7)2 =1.83×10−6 \begin{aligned} & f=\dfrac{{{K}_{a1}}\text{ }{{K}_{a2}}\text{ }{{K}_{a3}}}{{{K}_{a1}}\text{ }{{K}_{a2}}\text{ }{{K}_{a3}}+{{K}_{a1}}\text{ }{{K}_{a2}}\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ +}{{K}_{a1}}\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}{{\text{ }\\!\\!]\\!\\!\text{ }}^{\text{2}}}+{{\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{+}}}\text{ }\\!\\!]\\!\\!\text{ }}^{3}}} \\\ & =\dfrac{7.59\times {{10}^{-3}}\times 6.17\times {{10}^{-8}}\times 4.79\times {{10}^{-13}}}{7.59\times {{10}^{-3}}\times 6.17\times {{10}^{-8}}\times 4.79\times {{10}^{-13}}+7.59\times {{10}^{-3}}\times 6.17\times {{10}^{-8}}\times {{10}^{-7}}+7.59\times {{10}^{-3}}\times 6.17\times {{10}^{-8}}\times {{10}^{-7}}+7.59\times {{10}^{-3}}\times {{({{10}^{-7}})}^{2}}} \\\ & =1.83\times {{10}^{-6}} \\\ \end{aligned}

- Assume the solubility product of Zinc phosphate (Zn3(PO4)2Z{{n}_{3}}{{(P{{O}_{4}})}_{2}} ) is SM.

& [Z{{n}^{2+}}]=3SM \\\ & [PO_{4}^{3-}]=f\times 2S \\\ \end{aligned}$$ \- The formula of solubility product of zinc sulphate is $$\begin{aligned} & {{K}_{sp}}=[Z{{n}^{2+}}]{{[PO_{4}^{3-}]}^{2}} \\\ & 9.1\times {{10}^{-33}}={{(3S)}^{3}}\times {{(f\times 2S)}^{2}} \\\ & 9.1\times {{10}^{-33}}={{(3S)}^{3}}\times {{(1.83\times {{10}^{-6}}\times 2S)}^{2}} \\\ & S=3.0\times {{10}^{-5}}M \\\ \end{aligned}$$ \- Thus the solubility product of the $Z{{n}_{3}}{{(P{{O}_{4}})}_{2}}$ is $3.0\times {{10}^{-5}}M.$ \- But we have to calculate the concentration of the zinc. $$[Z{{n}^{2+}}] = 3S = 3\times 3.0\times {{10}^{-5}}=9\times {{10}^{-5}}M$$ \- Therefore the concentration of zinc is $9\times {{10}^{-5}}M$ . **So, the correct answer is “Option D”.** **Note:** ${{H}_{3}}P{{O}_{4}}$ is called phosphoric acid and it is a tribasic acid because it releases three hydrogen ions when dissolved in water. When phosphoric acid reacts with zinc it forms a zinc phosphate compound as the product.