The key here is determining the number of moles of NaOH so you can use the given stoichiometry to solve for how many grams of NaH2PO4 are needed. Looking at the reaction stoichiometry provided, you will need 1 moles of NaH2PO4 for every 2 moles of NaOH. Now we can work with what is given.
We start with 33.29 mL of 0.285 M NaOH. We need to convert this to moles so we can use the stoichiometric ratio. Molarity (M) has units of moles per liter, and we are given mL so we have to convert.
33.29 mL * 1 L ÷ 1000 mL = 0.03329 L NaOH solution
Now we multiply the molarity by the number of liters to get the number of moles. Again, we do this since the units of molarity are moles ÷ liters. If we multiply by liters, the liters cancel out and we are left with the unit moles.
0.285 M (or moles ÷ liters) * 0.03329 L = 0.00949 moles NaOH
This number may seem small, but remember we are only dealing with 33.29 mL which is smaller than a gulp of water so that isn’t out of the ordinary.
Now that we have moles of NaOH, we need to convert to moles of NaH2PO4. Remembering the stoichiometry from before, its 2 moles of NaOH per 1 mole of NaH2PO4. To get moles of NaH2PO4, divide the moles of NaOH by 2.
0.00949 moles NaOH * 1 mole NaH2PO4 ÷ 2 moles NaOH = 0.00474 moles NaH2PO4
We are almost done. We have converted mL of NaOH to moles of NaOH, and used the stoichiometry of the reaction to convert moles of NaOH to moles of NaH2PO4 needed for this reaction. The last step is using the number of moles of NaH2PO4 to determine the grams of NaH2PO4 required. This can be done by multiplying by the molecular weight, which has units of grams per mole. The molecular weight of NaH2PO4 is 119.98 g/mol. By multiplying by number of moles, the mol units will cancel out and all we are left with is grams.
0.00474 moles NaH2PO4 * 119.98 g ÷ moles = 0.569 g of NaH2PO4
Thus, our final answer would be 0.569 g of NaH2PO4.