Answer:
The reaction with sulphuric acid is given as:
\[Ba{{O}_{2}}.8{{H}_{2}}O+{{H}_{2}}S{{O}_{4}}(dil)\xrightarrow{{}}\]\[\underset{(White\,ppt.)}{\mathop{BaS{{O}_{4}}}}\,+{{H}_{2}}{{O}_{2}}+8{{H}_{2}}O\]
\[BaS{{O}_{4}}\] formed is slightly soluble in water. The \[B{{a}^{2+}}\] ions present in the solution accelerate or promote the decomposition of hydrogen peroxide to water and oxygen. In case phosphoric acid is used, then barium phosphate gets completely precipitated and the solution does not contain any \[B{{a}^{2+}}\] ions.
\[3Ba{{O}_{2}}.8{{H}_{2}}O+2{{H}_{3}}P{{O}_{4}}\xrightarrow[{}]{}\underset{(ppt.)}{\mathop{B{{a}_{3}}{{(P{{O}_{4}})}_{2}}}}\,+24{{H}_{2}}O+3{{H}_{2}}{{O}_{2}}\]
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