In a 0.5 M aqueous solution of H3PO4, which of the following relationships will exactly hold true?
(A) 0.5 = [H3PO4] + [H2PO2] + [HP0% 1+ [POZ]
(B) (H30*] = [H2PO4 ] + 2 [HPO 1 + 3[POZ] + [OH-]
(C) (H] = 3 [PO]
(D) [H30*1 = [H_PO,] + [HPOŽ ] + [PO?'] + [OH-] 1 / ]
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Answer:
(A)0.5=[H3PO4] + [H2PO2] + [HP0% 1+ [POZ]
Explanation:
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In a 0.5 M aqueous solution of H3PO4 is (A)0.5=[H3PO4] + [H2PO2] + [HP0% 1+ [POZ]
Explanation:
1. [H3PO4]+[H2PO4-]+[HPO42-]+[PO43-]
=0.5-x+x-y+y-2+7=0.5
2.H3PO4 is acid. It has a proton or hydrogen ion to donate in an aqueous solution. H3PO4 is acting as an Arrhenius acid and Bronsted-Lowry acid.
3.Strong acids are 100% ionized in solution. Weak acids are only slightly ionized. Phosphoric acid is stronger than acetic acid, and so is ionized to a greater extent.
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