50.00 mL of a 0.020 M solution of Ba(OH)₂, is added to 50.00 mL of a 0.060 M solution of HNO₃ - VCE - SSCE Chemistry - Question 6 - 2005 - Paper 1

For the HNO₃ solution:

$\text{Moles of HNO₃} = 0.060 \text{ M} \times 0.05000 \text{ L} = 0.00300 \text{ mol}$

Ba(OH)₂ reacts with HNO₃ according to the balanced equation:

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

From this, 1 mole of Ba(OH)₂ reacts with 2 moles of HNO₃, producing 2 moles of H⁺.

Considering the moles from above:

• Moles of OH⁻ contributed from Ba(OH)₂: 0.00200 mol (since Ba(OH)₂ provides 2 OH⁻)
• Moles of H⁺ from HNO₃: 0.00300 mol

Net moles of H⁺ after neutralization:

• Moles of OH⁻: 0.00200 mol neutralizes 0.00200 mol of H⁺
• Remaining H⁺: 0.00300 mol - 0.00200 mol = 0.00100 mol

The total volume of the solution after mixing is:

$50.00 ext{ mL} + 50.00 ext{ mL} = 100.00 ext{ mL} = 0.10000 ext{ L}$

Finally, calculate the concentration of H⁺:

$\text{Concentration of H⁺} = \frac{\text{Moles of H⁺}}{\text{Total Volume}} = \frac{0.00100 ext{ mol}}{0.10000 ext{ L}} = 0.0100 \text{ M}$

Thus, the hydrogen ion concentration is 0.010 M.