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Dimas [21]
3 years ago
7

The concentration of a NaOH solution was experimentally determined by dissolving 0.7816 grams KHP (204.2212 grams/mol) in 50.0 m

L of water and titrating the sample with 40.82 mL of the NaOH (40.000 grams/mol) solution. The concentration of the standard NaOH solution is _________.
Chemistry
1 answer:
REY [17]3 years ago
3 0

Answer:

The concentration of the standard NaOH solution is 0.094 moles/L.

Explanation:

In the titration, the equivalence point is defined as the point where the moles of NaOH (the titrant) and KHP (the analyte) are equal:

     moles of NaOH = moles of KHP

[NaOH]xV_{NaOH} = moles of KHP

[NaOH] = \frac{moles of KHP}{V_{NaOH}}

The V_{NaOH} is 40.82mL = 0.04082L and the moles of KHP are

0.7816g / 204.2212\frac{g}{mol} = 3.827x10^{-3} moles

Replacing at the first equation:

[NaOH] = \frac{3.827x10^{-3}moles}{0.04082L} = 0.094 moles/L

 

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