In a second titration with the same solution of NaOH as used in Question #1, Zac
ID: 1038325 • Letter: I
Question
In a second titration with the same solution of NaOH as used in Question #1, Zack used 0.31 g of KHP. Calculate the volume of the NaOH solution needed to neutralize this sample of KHP 2. 3. Write the neutralization equation which describes the reaction between the weak monoprotic acid such as formic acid, HCOOH (K, for the HCOOH is 1.8 x 104) and a strong base such as NaOH. 4. Ashley was given a 50.0 mL sample of 0.150 M solution of HCOOH. Using a burette, she transferred 25.0 mL of the weak acid into a 250 mL Erlenmeyer flask. Ashley then titrated the acid with a standardized solution of 0.0987 M NaOH. a) How many moles of the weak acid were added to the Erlenmeyer flask? b) How many moles of NaOH are required to neutralize (reach the equivalence point) the sample of weak acid? c) How many milliliters of the NaOH are required to neutralize the sample of weak acid? CHEM1515-YEAR 9017-2018 9-3Explanation / Answer
2) Need to know the molarity of NaOH to answer the question.
3) HCOOH is a monoprotic acid, i.e, has only one replaceable H+. Similarly, NaOH is a monobasic base, i.e, has only one replaceable OH-. H+ and OH- essentially combine to form H2O which remains undissociated. The neutralization reaction is given as
HCOOH (aq) + NaOH (aq) --------> NaCOOH (aq) + H2O (l)
4) As per the balanced stoichiometric equation in (3),
1 mole HCOOH = 1 mole NaOH.
(a) Ashley was give 50.0 mL of 0.150 M HCOOH and she transferred 25.0 mL of the weak acid to an Erlenmeyer flask.
Moles of HCOOH transferred to Erlenmeyer flask = (25.0 mL)*(1 L/1000 mL)*(0.150 M) = 0.00375 mole (ans).
(b) We have already noted that 1 mole NaOH neutralizes 1 mole HCOOH. Therefore, moles of NaOH required to neutralize 0.00375 mole HCOOH = (0.00375 mole HCOOH)*(1 mole NaOH/ 1 mole HCOOH) = 0.00375 mole (ans).
(c) Need to know the molarity of NaOH to find out the volume.
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