If 200. mL of 2.00 M H3PO4 solution is added to 600. mL of 2.00 M NaOH solution, the resulting solution will be ____ molar in Na3PO4.
you have .2*2 moles H2PO4
you have .6*2 moles of NaOH
balanced equation:
H3PO4 + 3NaOH>> Na3PO4 + H2O
so for each mole of acid, you need three times that amount of base.
.4moles acid, need 1.2 moles of base, but you have 1.2, which is magically just right. So, you should get .4 moles of trisodiumphosphate.
Molarity= .4 moles/(.800L+moleswater made)
moleswater made is .4 moles, or 18*.4grams, or 7.2 ml water or
molarity= .4moles/(.8072) then reduce that to three significant digits.
check my work.
To determine the molarity of Na3PO4 in the resulting solution, we need to perform a stoichiometric calculation based on the balanced chemical equation for the reaction between H3PO4 and NaOH:
H3PO4 + 3NaOH → Na3PO4 + 3H2O
First, we need to determine the amount of moles of H3PO4 and NaOH in their respective solutions:
Moles of H3PO4 = (volume of H3PO4 solution in liters) × (molarity of H3PO4)
= (200 mL ÷ 1000 mL/L) × (2.00 mol/L)
= 0.400 mol
Moles of NaOH = (volume of NaOH solution in liters) × (molarity of NaOH)
= (600 mL ÷ 1000 mL/L) × (2.00 mol/L)
= 1.200 mol
Since H3PO4 reacts with NaOH in a 1:3 stoichiometric ratio, the limiting reactant is H3PO4. This means that all of the moles of H3PO4 will react and only a third of that amount (in moles) of Na3PO4 will be produced.
Moles of Na3PO4 = (1/3) × moles of H3PO4
= (1/3) × 0.400 mol
= 0.133 mol
Next, we need to calculate the total volume of the resulting solution, which is the sum of the initial volumes of the two solutions:
Total volume of resulting solution = volume of H3PO4 solution + volume of NaOH solution
= 200 mL + 600 mL
= 800 mL
Finally, we can calculate the molarity of Na3PO4 in the resulting solution by dividing moles of Na3PO4 by the volume of the resulting solution in liters:
Molarity of Na3PO4 = (moles of Na3PO4) ÷ (volume of resulting solution in liters)
= 0.133 mol ÷ (800 mL ÷ 1000 mL/L)
= 0.166 M
Therefore, the resulting solution will be approximately 0.166 M in Na3PO4.
To find the resulting molarity of Na3PO4, you need to use the concept of stoichiometry and the balanced chemical equation that represents the reaction between H3PO4 and NaOH.
The balanced chemical equation for the reaction is:
H3PO4 + 3NaOH -> Na3PO4 + 3H2O
From the equation, you can see that for every 1 mole of H3PO4, you will form 1 mole of Na3PO4. This means the moles of Na3PO4 formed will be equal to the moles of H3PO4 added.
To find the moles of H3PO4 in the solution, you can use the formula:
moles = molarity × volume (in liters)
First, convert the volume of the H3PO4 solution from milliliters (mL) to liters (L). Since there are 200. mL of H3PO4 solution, the volume in liters is 200. mL ÷ 1000 = 0.200 L.
Now calculate the moles of H3PO4:
moles of H3PO4 = molarity of H3PO4 × volume of H3PO4 solution
= 2.00 M × 0.200 L
= 0.400 moles
Since the balanced equation shows a 1:1 mole ratio between H3PO4 and Na3PO4, this means that 0.400 moles of Na3PO4 will be formed.
Finally, calculate the molarity of Na3PO4 by dividing the moles of Na3PO4 by the total volume of the resulting solution (the sum of volumes of H3PO4 and NaOH solutions):
Molarity of Na3PO4 = Moles of Na3PO4 ÷ Total volume of solution
The total volume of the solution is the sum of the volumes of the H3PO4 and NaOH solutions, which is:
0.200 L (H3PO4 solution) + 0.600 L (NaOH solution) = 0.800 L
Now calculate the molarity of Na3PO4:
Molarity of Na3PO4 = 0.400 moles ÷ 0.800 L
= 0.500 M
Therefore, the resulting solution will be 0.500 M in Na3PO4.