Part A. If 31.7mL of 0.230M KOH is required to completely neutralize 30.0mL of a

Question

Part A.

If 31.7mL of 0.230M KOH is required to completely neutralize 30.0mL of a HC2H3O2 solution, what is the molarity of the acetic acid solution?
HC2H3O2(aq) + KOH(aq) ? KC2H3O2(aq) + H2O(l)

Part B.

What is the [H3O+] in a solution that contains 1.72g HNO3 in 0.540L of solution?

Part C.

A 12.0mL sample of vinegar, which is an aqueous solution of acetic acid, HC2H3O2, requires 17.5mL of 0.300M NaOH to reach the endpoint in a titration.

HC2H3O2(aq) + NaOH(aq) ?
NaC2H3O2(aq) + H2O(l)

What is the molarity of the acetic acid solution?

Explanation / Answer

If 31.7mL of 0.230M KOH is required to completely neutralize 30.0mL of a HC2H3O2 solution, what is the molarity of the acetic acid solution?
HC2H3O2(aq) + KOH(aq) ---->KC2H3O2(aq) + H2O(l)

Given : Volume of KOH = 31.7 mL , Molarity = 0.230 M

Volume of HC2H3O2 = 30.0 mL

Solution :

HC2H3O2(aq) + KOH(aq) --> KC2H3O2(aq) + H2O(l)
From the reaction stoichiometry we say that mol ratio of HC2H3O2(aq): KOH is 1 : 1

Lets find moles of KOH from given volume and molarity

mol KOH = Molarity * Volume in L

= 0.230 M * 0.0317 L = 0.00729 mol KOH

Lets calculate moles of HC2H3O2

HC2H3O2 = mol KOH * 1mol HC2H3O2/ 1 mol KOH

= 0.00729 mol HC2H3O2

We know molarity = mol / Volume in L

Molarity of HC2H3O2 = 0.00729 mol HC2H3O3 / 0.030 L

=0.243 M

What is the [H3O+] in a solution that contains 1.72g HNO3 in 0.540L of solution?

Given :

Mass of HNO3 = 1.72 g

Volume of HNO3 = 0.540 L

Molarity [H3O+] =[HNO3]

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects