A student uses a weak monoprotic carboxylic acid with a molecular weight of 204.

Question

A student uses a weak monoprotic carboxylic acid with a molecular weight of 204.23 g/mol to standardize an NaOH solution. He dissolved 0.500 g of the weak acid in 100.0 mL of DI water and adds phcnolphthalein indicator. This solution is then titrated with NaOH until the solution turns pink. The initial reading of the buret containing the NaOH was 50.0 mL. How many moles of weak acid were in the flask? How many moles of NaOH were used to reach the equivalence point? The final buret reading was 36.4 mL. Calculate the molarity of the NaOH.

Explanation / Answer

acid moles = mass / Molar mass = 0.5/204.23 =0.002448

A) Number of moles of weak acid = 0.002448

B) at equivalence point acid moles = base moles   since acid and base react in 1:1 ratio ,

HA + NaOH ---> NaA + H2O   is acid base reaction , where HA is weak monoprotic acid representation

hence naOH moles = 0.002448

C) Initial buret reading = 50 ml , final is 36.4 ml ,

NaOH volume used = 50-36.4 = 13.6 ml = 13.6/1000 L = 0.0136 L

we have formula Molarity = moles / volume in L of solution

hence Molarity of NaOH = ( moles of NaOH ) / ( vol of NaOH used)

                                 = ( 0.002448 /0.0136)

                          = 0.18 M

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