A 30.0 mL sample of phosphoric acid is titrated with 18.0 mL of 3.0 M NaOH solution. What is the molarity of the phosphoric acid?

Phosphoric acid has three H ions; therefore, technically at least, it will have three inflections in the curve. I assume you mean for ALL of the H3PO4 to be neutralized.

3NaOH + H3PO4 ==> Na3PO4 + H2O

moles NaOH = M x L
moles H3PO4 = moles NaOH x 1/3
M H3PO4 = moles H3PO4/L H3PO4

To find the molarity of the phosphoric acid, we can use the concept of stoichiometry. The balanced chemical equation for the reaction between phosphoric acid (H3PO4) and sodium hydroxide (NaOH) is:

H3PO4 + 3NaOH → Na3PO4 + 3H2O

From the equation, we can see that 1 mole of H3PO4 reacts with 3 moles of NaOH. Since we know the volume and concentration of NaOH used in the titration, we can calculate the number of moles of NaOH:

Number of moles of NaOH = Volume of NaOH (in liters) × Concentration of NaOH (in moles per liter)

Converting the given volume of NaOH to liters:

18.0 mL ÷ 1000 = 0.018 L

Then, plugging in the values to find the moles of NaOH:

Number of moles of NaOH = 0.018 L × 3.0 mol/L = 0.054 mol

Since the ratio between H3PO4 and NaOH is 1:3, the moles of H3PO4 will be one-third of the moles of NaOH:

Number of moles of H3PO4 = 1/3 × 0.054 mol = 0.018 mol

Now, we can calculate the molarity of the phosphoric acid:

Molarity (M) = Number of moles of H3PO4 / Volume of phosphoric acid (in liters)

Converting the given volume of phosphoric acid to liters:

30.0 mL ÷ 1000 = 0.03 L

Then, plugging in the values to find the molarity:

Molarity (M) = 0.018 mol / 0.03 L = 0.6 M

Therefore, the molarity of the phosphoric acid is 0.6 M.