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0.095 g of an unknown diprotic acid is titrated with 0.095 M NaOH. The first equivalence point occurred in the titration at a volume of 6.70 mL of NaOH added; the second equivalence point occurred at a volume of 13.40 mL of NaOH added. How many moles of NaOH were used to reach the first equivalence point in this diprotic acid titration

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Answer:

Explanation:

The first equivalence point occurred in the titration at a volume of 6.70 mL of NaOH added . The molarity of NaOH is .095 M .

6.70 mL of NaOH = .00067 L of NaOH .

.00067 L of .095 M NaOH will contain .00067 x .095 moles of NaOH .

= 6.365 x 10⁻⁵ moles .

Moles of NaOH  used to reach the first equivalence point = 6.365 x 10⁻⁵ moles.

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