Question #02722

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Doc048
May 21, 2017

f.wt. = 125 g/mole

Explanation:

Assume the following acid/base rxn

HA + NaOH => NaA + HOH
Rxn is 1:1 mole ratio of HA:HOH which means that moles of acid will be completely neutralized by an equal number of moles of base. That is ...

moles acid neutralized = moles of base used

moles = mass(g)/formula wt and in solution, moles = Molarity x Volume in Liters

Given:
#"moles" of HA = 0.25 "grams" / "f.wt. of acid"#
moles of NaOH = #((0.20 mol)/(dL)^3)(0.010Liters)# = #((0.20 mol)/L)(0.010L)# = #0.002 "mole"#

using moles acid = moles base

#(0.25 grams)/(f.wt acid)# = #(0.002"mole")#

f.wt acid = #(0.25 g)/((0.002"mole")# = #125 g/"mol"#

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