When 22.63 mL of aqueous NaOH was added to 1.136 g of cyclohexylaminoethanesulfonic acid (FM 207.29,...

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When 22.63 mL of aqueous NaOH was added to 1.136 g ofcyclohexylaminoethanesulfonic acid (FM 207.29, pK_a =9.39, structure in the table) dissolved in 41.37 mL of water, thepH was 9.24. Calculate the molarity of the NaOH.

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4.1 Ratings (428 Votes)

number of moles of cyclohexylaminoethanesulfonic acid = (1.136 / 207.29)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 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)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.004435 M

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