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V125BC
4 months ago
9

What volume (ml) of a 0.2450 m koh(aq) solution is required to completely neutralize 55.25 ml of a 0.5440 m h3po4(aq) solution?

Chemistry
1 answer:
VMariaS [2.9K]4 months ago
5 0
<span>Answer: The answer varies based on the type of acid following "0.5440 M H...". For a monoprotic acid such as HCl, the calculation proceeds as follows: (55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) = 122.7 mL KOH In the case of a diprotic acid like H2SO4, it works out to: (55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) = 245.4 mL KOH For a triprotic acid such as H3PO4, it is: (55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) = 368.0 mL KOH</span>
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