Question

FAcid-Base Titrations ③ 4088 | Review Constants Periodic Table You are given a solution of HCOOH (formic acid) with an approximate concentration of 0.20 M and you will titrate this with a 0.1105 M NaOH. If you add 20.00 mL of HCOOH to the beaker before titrating, approximately what volume of NaOH will be required to reach the end point? View Available Hint(s) O 20.0 mL O 11.1 ml 72.4 mL O 36.2 mL Submit Part B

Answer 1

Here is the attachment of the answer. Hope it'll help out. Please do give positive rating of you are satisfied with the answer. Given,

molarity of NaOH(M1) =0.1105 Mol/L

Volume of NaOH (V1)= ? mL

Volume of formic acid(HCOOH) (V2)=20.00 mL

Molarity of formic acid (M2)=0.20 Mol/L

METHOD-1

M1V1(NaOH)= M2V2 (HCOOH)

V1 = M2V2/M1

V1= 0.20 M *20.00mL/ 0.1105 M

Answer::::::Volume of NaOH required for neutrilisation of HCOOH = 36.2 mL

METHOD-2

moles of HCOOH in Beaker= Molarity/ Volume taken in the Beaker

Mole of HCOOH= 0.20 mol.L-*20 L/1000

mole of HCOOH= 0.004 mol

now, one mole HCOOH reacts with 1 mole NaOH. So mole of NaOH required are also 0.004 mol.

NaOH +HCOOH =====> HCOO-Na+ + H2O

Molarity = Mole/ L

Volume(L) = Mole/ Molarity

Volume of NaOH = 0.004(mol)/0.1105 (mol.L-)

Volume of NaOH = 0.0362 L* 1000= 36.2 m L