Question

What is the final solution pH after 60 mL 0.05 M Ba(OH)2 is mixed with 25 mL 0.20 M HNO3?

Answer 1

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68538 Chemistry, Other

What is the final solution pH after 60 mL 0.05 M Ba(OH)₂ is mixed with 25 mL 0.20 M HNO₃?

Answer:

$$\mathrm{Ba(OH)}_2 + 2\mathrm{HNO}_3 = \mathrm{Ba(NO}_3)_2 + 2\mathrm{H}_2\mathrm{O}$$

$\mathrm{Ba(OH)}_2$ is a strong alkali. $\mathrm{HNO}_3$ is a strong acid.

In order to determine pH, we have to determine limiting reagent.

According to equation, $n(\mathrm{Ba(OH)}_2) = 2 \cdot n(\mathrm{HNO}_3)$

$$C_M = n/V \quad n = C_M \cdot V$$ $$n(\mathrm{Ba(OH)}_2) = 0.05 \cdot 60 = 3 \text{ mmol}$$ $$n(\mathrm{HNO}_3) = 0.2 \cdot 25 = 5 \text{ mmol} \quad \text{But it must be: } 2 \cdot n(\mathrm{Ba(OH)}_2) = 3 \cdot 2 = 6 \text{ mmol}$$

Therefore, $\mathrm{HNO}_3$ is the limiting reagent. There will be an excess of $\mathrm{Ba(OH)}_2$ of $(3-5/2=0.5 \text{ mmol})$ with an anticipated basic pH.

$$\mathrm{pOH} = -\lg[0.0005] = 3.3$$ $$\mathrm{pH} + \mathrm{pOH} = 14$$ $$\mathrm{pH} = 14-3.3 = 10.7$$