Use the following balanced equation.
3Sr(OH)2 + 2H3PO4 --> 6H20 + Sr3(PO4)2
20 mL of H3PO4 is required to react with 40 mL of a 0.15 M Sr(OH)2 solution. What is the molarity of the H3PO4 solution?
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To find the molarity of the H3PO4 solution, we need to use the balanced equation to establish a stoichiometric relationship between the two solutions.
According to the balanced equation, the ratio of Sr(OH)2 to H3PO4 is 3:2. This means that for every 3 moles of Sr(OH)2, we need 2 moles of H3PO4.
First, let's convert the volume of the Sr(OH)2 solution (40 mL) into moles. Since we know the concentration of Sr(OH)2 is 0.15 M (moles per liter), we can use the formula:
moles = concentration x volume (in liters)
moles of Sr(OH)2 = 0.15 M x (40 mL / 1000 mL/ L) = 0.006 moles
Now, using the stoichiometric ratio, we can determine the moles of H3PO4 required for the reaction. Since the ratio of Sr(OH)2 to H3PO4 is 3:2, we can set up the following proportion:
3 moles Sr(OH)2 / 2 moles H3PO4 = 0.006 moles Sr(OH)2 / x moles H3PO4
Solving for x (the moles of H3PO4):
x = (2 moles H3PO4 / 3 moles Sr(OH)2) * 0.006 moles Sr(OH)2 = 0.004 moles H3PO4
Finally, we can find the molarity of the H3PO4 solution by dividing the moles of H3PO4 by its volume in liters:
Molarity = moles / volume (in liters)
We were given the volume of the H3PO4 solution as 20 mL, so converting it to liters:
Molarity = 0.004 moles H3PO4 / (20 mL / 1000 mL/L) = 0.2 M
Therefore, the molarity of the H3PO4 solution is 0.2 M.