Question

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)

Answer 1

Answer:

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

Answer 2

Answer:

$M_{base}=0.957M$

Explanation:

Hello,

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

- The normality of phosphoric acid (triporitic) is:

$N_{acid}=0.314\frac{mol}{L}*\frac{3eq-g}{1mol}=0.942N$

Thus, the normality of barium hydroxide turns out:

$N_{base}=\frac{N_{acid}V_{acid}}{V_{base}}=\frac{0.942N*32.7mL}{16.1mL} =1.91N$

Finally, the molarity:

$M_{base}=1.91\frac{eq-g}{L}*\frac{1mol}{2eq-g}=0.957M$

Best regards.