If 33.7 ml of 0.210 M KOH is required to completely neutralize 27.0 ml of a HC2H3O2 solution, what is the molarity of the acetic acid solution?
HC2H3O2(aq)+ KOH(aq)-->KC2H3O2(aq)+H2O(l)
mols KOH = M x L = ?
mols acetic acid = mols KOH (from the coefficients in the balanced equation.)
M acid = mols acid/L acid. You know mols and L, solve for M.
To find the molarity of the acetic acid solution (HC2H3O2), you'll need to use the equation for the neutralization reaction and the given information.
The equation for the reaction is:
HC2H3O2(aq) + KOH(aq) → KC2H3O2(aq) + H2O(l)
From the equation, you can see that the mole ratio between HC2H3O2 and KOH is 1:1. This means that for every 1 mole of HC2H3O2, 1 mole of KOH is required to neutralize it.
Given that 33.7 ml of 0.210 M KOH is required to completely neutralize the HC2H3O2 solution, we can use this information to find the moles of KOH used.
Molarity (M) is defined as moles of solute per liter of solution. Since the volume of KOH used is given in milliliters, we need to convert it to liters by dividing by 1000:
33.7 ml / 1000 = 0.0337 L
Now we can calculate the moles of KOH:
moles KOH = Molarity (M) × volume (L)
moles KOH = 0.210 M × 0.0337 L = 0.007077 moles KOH
Since the mole ratio between HC2H3O2 and KOH is 1:1, the moles of HC2H3O2 used will also be 0.007077 moles.
The molarity of the acetic acid solution can now be calculated using the following equation:
Molarity (M) = moles of solute / volume of solution (in liters)
Given that the volume of the acetic acid solution is 27.0 ml, we need to convert it to liters by dividing by 1000:
27.0 ml / 1000 = 0.027 L
Now we can calculate the molarity of the acetic acid solution:
Molarity of acetic acid = 0.007077 moles HC2H3O2 / 0.027 L = 0.262 M
Therefore, the molarity of the acetic acid solution is 0.262 M.