Question

Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the excess H1 or OH2 ions left in this solution?

Answer 1

Answer:

Concentration of unreacted hydroxide ions in the final solution is 0.02 mol/L.

Explanation:

$Molarity=\frac{moles}{\text {Volume in L}}$

Moles of HCl

=$Molarity\times {\text {Volume in L}}=0.250 M\times 0.075L=0.01875 moles$

1 mole of HCl give = 1 mole of $H^+$

Thus 0.01875 moles of HCl give = 0.01875 mole of $H^+$

Moles of $Ba(OH)_2$ :

$=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012375 mol$

According to stoichiometry:

1 mole of Ba(OH)_2 gives = 2 moles of $OH^-$:

Thus 0.012 moles of $Ba(OH)_2$ give = 2 \times 0.012375 =0.02475 moles of $OH^-$

According to neutralization reaction, 1 mole of $H^+$ neutralizes 1 mol $OH^-$.

0.01875 mol of $H^+$ will neutralize 0.01875 mol $OH^-$.

Moles of hydroxide ion left unreacted :

= 0.02475 mol - 0.01875 mol = 0.006 mol

Concentration of unreacted hydroxide ions in the final solution:

$\frac{0.006 mol}{0.225 L+0.075 L}=0.02 mol/L$

Answer 2

Answer:  The concentration of excess $[OH^-]$ in solution is 0.017 M.

Explanation:

1. $Molarity=\frac{moles}{\text {Volume in L}}$

moles of $HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles$

1 mole of $HCl$ give = 1 mole of $H^+$

Thus 0.019 moles of $HCl$ give = 0.019 mole of $H^+$

2. moles of $Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles$

According to stoichiometry:

1 mole of $Ba(OH)_2$ gives = 2 moles of $OH^-$

Thus 0.012 moles of $Ba(OH)_2$ give = $2 \times 0.012=0.024$ moles of $OH^-$

$H^++OH^-\rightarrow H_2O$

As 1 mole of $H^+$ neutralize 1 mole of $OH^-$

0.019 mole of $H^+$ will neutralize 0.019 mole of $OH^-$

Thus (0.024-0.019)= 0.005 moles of $OH^-$ will be left.

$[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M$

Thus molarity of $[OH^-]$ in solution is 0.017 M.