Formic Acid (1.0M)-Sodium Formate (68.01 g/mol) Paragraph from page 4: Prepare 0

Question

Formic Acid (1.0M)-Sodium Formate (68.01 g/mol)
Paragraph from page 4:
Prepare 0.10 M solutions with volumes of 100mL of both the assigned acid and it's base. For most of the acids this will involve diluting a 1.0M solution of the acid. For most of the bases this will involve adding the appropriate mass of the base to a 100mL volumetric flask and diluting to mark.

m Formate idenoty of Assigned AcidBase Conjugate Par Formie acid cancentration of stock Solution of Add (or Base in oase of Ammonay Actual quantity of acid used tity of base used Prior to Class Calculations: Calculate the volume and mass required to make o.10 M solutions of your assigned acid and conjugate base see page 4 for concentrations and

Explanation / Answer

To prepare one L of 1M stock solution of both acid and conjugate base

1L of 1M formic acid stock solution

Molar mass of formic acid HCOOH = 46g/mol

molarity = number of moles /V(L)

= (46g/46g/mol) /1L = 1 M

Thus dissolving 46g of formic acid in 1 L of solution gives 1M solution.

!L of 1M sodium formate solution

molar mass of sodium formate = 68.01g/mol

Molarity = (68.01g/68.01g/mol ) /1L

= 1M

Thus dissolving 68.01 g in 1 L solution gives 1M sodium formate.

Part II

preparation of 0.1M solution of acid and conjugate base

V x1M = 100mL x 0.1M

thus V = 10mL

Thus diluting 10mL of 1M stock solution into 100mL gives 0.1M solution.

(same for both acid and sodium formate)

Mass required to make 100mL of 0.1 M solution

For acid solution

mass = molarity x molar mass x V(L)

= 0.1mol/L x46g/mol x0.1L

= 0.46g

similarly mass of sodium formate in 100mL of 0.1M solution

= 0.1mol/L x 68.01g/mol x 0.1L

= 0.68g

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