25.14 ml of 0.2220 m phosphoric acid is neutralized by 38.77 ml of barium hydroxide solution. determine the…

Explanation:

Step1: Write balanced reaction

$$2\text{H}_3\text{PO}_4 + 3\text{Ba(OH)}_2 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6\text{H}_2\text{O}$$

Step2: Moles of $\text{H}_3\text{PO}_4$

Moles = Molarity × Volume (L) $$n_{\text{H}_3\text{PO}_4} = 0.2220\ \text{mol/L} \times 0.02514\ \text{L} = 0.00558108\ \text{mol}$$

Step3: Mole ratio for $\text{Ba(OH)}_2$

From reaction, $\frac{n_{\text{Ba(OH)}2}}{n{\text{H}_3\text{PO}4}} = \frac{3}{2}$ $$n{\text{Ba(OH)}_2} = 0.00558108\ \text{mol} \times \frac{3}{2} = 0.00837162\ \text{mol}$$

Step4: Molarity of $\text{Ba(OH)}_2$

Molarity = $\frac{\text{Moles}}{\text{Volume (L)}}$ $$M_{\text{Ba(OH)}_2} = \frac{0.00837162\ \text{mol}}{0.03877\ \text{L}}$$

Answer:

$0.2159\ \text{M}$