A 25.00 mL sample of H3PO4(aq) requires 31.05 mL of 0.2410 M KOH for titration to the second equivalence point.
What is the molarity of the H3PO4(aq)?
WEll, you have titrated two of the protons..
.025*2Ma=.03105*.2410
solve for Molarity of the acid, Ma
To find the molarity of H3PO4(aq), we can use the equation for neutralization reactions:
H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(l)
From the balanced equation, we can see that the ratio of H3PO4 to KOH is 1:3.
Given the volume and molarity of KOH used in the titration (31.05 mL of 0.2410 M KOH), we can set up a proportion to find the moles of KOH used:
(0.2410 mol KOH / 1000 mL) x 31.05 mL = X mol KOH
Next, we use the stoichiometry of the neutralization reaction to find the moles of H3PO4:
(1 mol H3PO4 / 3 mol KOH) x X mol KOH = Y mol H3PO4
To find the molarity of H3PO4, we divide the moles of H3PO4 by the volume of H3PO4 used (25.00 mL = 0.02500 L):
Y mol H3PO4 / 0.02500 L = Molarity of H3PO4
Now let's calculate it step by step:
Step 1: Calculate the moles of KOH used in the titration.
(0.2410 mol KOH / 1000 mL) x 31.05 mL = 0.007493 mol KOH
Step 2: Calculate the moles of H3PO4.
(1 mol H3PO4 / 3 mol KOH) x 0.007493 mol KOH = 0.002498 mol H3PO4
Step 3: Calculate the molarity of H3PO4.
0.002498 mol H3PO4 / 0.02500 L = 0.09992 M
Therefore, the molarity of the H3PO4(aq) is approximately 0.09992 M.