A 16.1 mL sample of Ba(OH)2 is titrated with H3PO4. If 32.7 mL of 0.314 M H3PO4 is needed to reach the endpoint, what is the concentration (M) of the Ba(OH)2 solution? 3 Ba(OH)2(aq) + 2 H3PO4(aq) → Ba3(PO4)2(aq) + 6 H2O(l)

Explanation:

Hello,

This problem could be solved in terms of normality as follows:

- The normality of phosphoric acid (triporitic) is:

Thus, the normality of barium hydroxide turns out:

Finally, the molarity:

Best regards.

0.963 M

Explanation:

Let's consider the following neutralization reaction.

3 Ba(OH)₂(aq) + 2 H₃PO₄(aq) → Ba₃(PO₄)₂(aq) + 6 H₂O(l)

32.7 mL of 0.314 M H₃PO₄ is needed to reach the endpoint. The reacting moles of H₃PO₄ are:

0.0327 L × 0.314 mol/L = 0.0103 mol

The molar ratio of Ba(OH)₂ to H₃PO₄ is 3:2. The reacting moles of Ba(OH)₂ are 3/2 × 0.0103 mol = 0.0155 mol

0.0155 moles of Ba(OH)₂ are in 16.1 mL. The molarity of Ba(OH)₂ is:

M = 0.0155 mol / 0.0161 L = 0.963 M

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.