A volume of 32.45 mL of 0.122 M NaOH was required to titration 4.89 g of vinegar to a phenolphthalein end point. How many moles of NaOH were added?

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A volume of 32.45 mL of 0.122 M NaOH was required to titration 4.89 g of vinegar to a phenolphthalein end point. How many moles of NaOH were added?

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Answer 1 · 2015-11-12T15:10:59

Concentration (#c#) is always moles (or sometimes mass) per unit volume. If #c# #=# #n/V#, then #n# (molar quantity) #=# volume #(V)# #xx# concentration #n/V#.

Explanation:

#n# #=# #Vxxc# #=# #32.45# #xx10^(-3)cancelL# #xx# #0.122# #mol*cancelL^-1#

#= 3.96xx10^(-3) mol# #NaOH#

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A volume of 32.45 mL of 0.122 M NaOH was required to titration 4.89 g of vinegar to a phenolphthalein end point. How many moles of NaOH were added?

1 Answer

Explanation:

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