Question

3. What is the molarity of a Sr(OH)2 solution if 40 mL is required to neutralize 70 mL of

a 8.5 M solution of H3PO4?

Answer 1

In this problem, we can use the concept of acid-base titration to determine the molarity of the Sr(OH)2 solution. The balanced chemical equation for the reaction between H3PO4 and Sr(OH)2 is:

3H3PO4 + Sr(OH)2 → Sr(H2PO4)2 + 2H2O

From the equation, we can see that 3 moles of H3PO4 react with 1 mole of Sr(OH)2. Therefore, the number of moles of H3PO4 in the solution is:

moles of H3PO4 = Molarity × Volume = 8.5 M × 0.070 L = 0.595 moles

Since 3 moles of H3PO4 react with 1 mole of Sr(OH)2, the number of moles of Sr(OH)2 in the solution is:

moles of Sr(OH)2 = (1/3) × 0.595 moles = 0.1983 moles

The volume of the Sr(OH)2 solution is 40 mL, or 0.040 L. Therefore, the molarity of the Sr(OH)2 solution is:

Molarity = moles of Sr(OH)2 / Volume of solution = 0.1983 moles / 0.040 L = 4.96 M

Therefore, the molarity of the Sr(OH)2 solution is 4.96 M.