Question

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

Answer 1

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:

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