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?

Answer 1

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#.

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

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

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Answer 2

Peyton Adams

To find the number of moles of NaOH added, you can use the formula:

[ \text{moles} = \text{Molarity} \times \text{Volume (L)} ]

Given:

Molarity ((M)) of NaOH = 0.122 (M)
Volume of NaOH used = 32.45 (mL) (which is 0.03245 (L))

[ \text{moles} = 0.122 \times 0.03245 = 0.003968 \text{ moles of NaOH} ]

Therefore, 0.003968 moles of NaOH were added.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.