A solution of 0.316 M KOH is used to neutralize 16.0 mL H3PO4 solution
H3PO4(aq) + 3KOH(aq) �¨ K3PO4(aq) + 3H2O(l)
If 30.3 mL KOH solution is required to reach the endpoint, what is the molarity of the H3PO4 solution?
Molarity = ____________
mols KOH = M x L = ?
Using the coefficients in the balanced equation, convert mols KOH to mols H3PO4.
Then M H3PO4 = mols H3PO4/L H3PO4.
To find the molarity of the H3PO4 solution, we need to use the stoichiometry of the balanced equation and the volume and molarity of the KOH solution used.
First, let's calculate the number of moles of KOH used. The molarity (M) of KOH solution is given as 0.316 M, and the volume (V) used is 30.3 mL. We need to convert the volume to liters:
V(KOH) = 30.3 mL = 30.3 mL * (1 L / 1000 mL) = 0.0303 L
Now we can calculate the moles of KOH:
moles(KOH) = Molarity * Volume = 0.316 M * 0.0303 L = 0.00957 moles
According to the balanced equation, the stoichiometric ratio between H3PO4 and KOH is 1:3. This means that for every 1 mole of H3PO4, we need 3 moles of KOH.
Therefore, the moles of H3PO4 in the solution is 1/3 of the moles of KOH used:
moles(H3PO4) = 1/3 * moles(KOH) = 1/3 * 0.00957 moles = 0.00319 moles
Finally, we can calculate the molarity of the H3PO4 solution. The molarity (M) is defined as moles of solute divided by the volume of the solution in liters. The volume is given as 16.0 mL, which needs to be converted to liters:
V(H3PO4) = 16.0 mL = 16.0 mL * (1 L / 1000 mL) = 0.016 L
Now, we can calculate the molarity:
Molarity(H3PO4) = moles(H3PO4) / Volume(H3PO4) = 0.00319 moles / 0.016 L = 0.1994 M
Therefore, the molarity of the H3PO4 solution is 0.1994 M.