In: Chemistry
When 22.63 mL of aqueous NaOH was added to 1.136 g of
cyclohexylaminoethanesulfonic acid (FM 207.29, pKa =
9.39, structure in the table) dissolved in 41.37 mL of water, the
pH was 9.24. Calculate the molarity of the NaOH.
number of moles of cyclohexylaminoethanesulfonic acid = (1.136 / 207.29)
= 5.48 * 10^-3 moles
pH = pKa + log((salt) / (acid))
The formation of salt depends upon the base because it can dissociate compltely in ions.
9.24 = 9.39 + log( (x * 22.63) / ((5.48*10^-3) - (x * 22.63)))
number of mole of NaOH = 1.0037 * 10^-4
concentration of NaOH = (1.0037 * 10^-4) * (1000 / 22.63)
= 0.004435 M