if 11.65 ml of 0.595 m koh is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid?

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19-12-2022
Chemistry

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if 11.65 ml of 0.595 m koh is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid?

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SidHussain SidHussain · Answer 1 · 2022-12-22T14:20:09-05:00

The concentration of the unknown acid is 6.93175 divided by the volume of the unknown acid in mL and the unit is M (mole/liter)

The formula for volumetric analysis can be given as follows;
[tex]\frac{C_{a}V_{a} }{C_{b}V_{b} } = \frac{N_{a} }{N_{b} }[/tex]

At the equivalence point, [tex]N_{a} = N_{b}[/tex]

Therefore, [tex]C_{a} V_{a} = C_{b} V_{b}[/tex]

[tex]C_{a}[/tex] (concentration of unknown acid) = [tex]\frac{C_{b}V_{b} }{V_{a} }[/tex]

[tex]C_{a}[/tex] represents the concentration of unknown acid in mL

[tex]C_{b}[/tex] (concentration of KOH) = 0.595M

[tex]V_{b}[/tex] (volume of KOH) = 11.65 mL

[tex]C_{a}[/tex] = 11.65 × 0.595 / [tex]V_{a}[/tex] = (6.93175 / [tex]V_{a}[/tex]) M

Hence, the concentration of the unknown acid is calculated to be 6.93175 divided by the volume of the unknown acid in mL if 11.65 ml of 0.595 KOH is required for titration.

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