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

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

Step-by-step 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] = (moles of KHP)/(V_(NaOH))`

The
`V_(NaOH)` is 40.82mL = 0.04082L and the moles of KHP are

`0.7816g / 204.2212(g)/(mol) = 3.827x10^(-3) moles`

Replacing at the first equation:

`[NaOH] = (3.827x10^(-3)moles)/(0.04082L) = 0.094 moles/L`