Final answer:
The final concentration of hydroxide ions after Ba(OH)2 reacts with H3PO4 is 0.1283 M. Stoichiometric calculations are used to determine the amount of Ba(OH)2 that reacts and the amount that remains unreacted, which then allows for the hydroxide ion concentration to be calculated.
Step-by-step explanation:
The final concentration of hydroxide ions when 2.00 L of 0.100 M Ba(OH)2(aq) is combined with 1.00 L of 5.00×10-3M H3PO4(aq) will depend on the reaction between barium hydroxide, a strong base, and phosphoric acid, a weak acid. The balanced chemical equation for the reaction between barium hydroxide and phosphoric acid is:
3Ba(OH)2 + 2H3PO4 → Ba3(PO4)2 + 6H2O
Initially, we have 0.200 moles of Ba(OH)2 (0.100 M × 2.00 L) and 0.005 moles of H3PO4 (5.00×10-3 M × 1.00 L). Since phosphoric acid is the limiting reactant, it will completely react with Ba(OH)2 to form 0.005 moles of Ba3(PO4)2. To find out the amount of Ba(OH)2 left after the reaction, we can perform stoichiometric calculations but we would need one mole of H3PO4 for every three moles of Ba(OH)2, therefore, 0.005 / 2 * 3 = 0.0075 moles of Ba(OH)2 will react. By subtracting the used moles from the initial moles of Ba(OH)2, we get 0.1925 moles of Ba(OH)2 that did not react. Each mole of Ba(OH)2 produces two moles of hydroxide ions, so the remaining hydroxide ion concentration is (0.1925 moles × 2) / (2.00 L + 1.00 L) = 0.1283 M.