Question

Hydrochloric acid (28.5mL 0.109M) is added to 298.0mL 0.0830M solution. What is the concentration of the excess OH− ions left in this solution? Assume that the volumes are additive.

Answer 1

To calculate the excess OH- concentration, we determine that NaOH is in excess after reaction with HCl. We then find the concentration by dividing the moles of excess NaOH by the total volume of the mixture, resulting in a final concentration of 0.0662 M for OH- ions.

Step-by-step explanation:

To find the concentration of excess OH− ions after hydrochloric acid (HCl) is added to an NaOH solution, we must first consider the reaction between HCl and NaOH:

• HCl + NaOH → NaCl + H2O

This reaction will neutralize an equivalent amount of acid and base. With 28.5 mL of 0.109 M HCl and 298.0 mL of 0.0830 M NaOH, we can calculate the moles of HCl and NaOH added:

• Moles of HCl = Molarity HCl × Volume HCl (in liters)
• Moles of HCl = 0.109 mol/L × 0.0285 L = 0.0031075 mol
• Moles of NaOH = Molarity NaOH × Volume NaOH (in liters)
• Moles of NaOH = 0.0830 mol/L × 0.2980 L = 0.024734 mol

Now we can determine which reactant is in excess. Since more moles of NaOH are present than moles of HCl, NaOH is in excess:

• Excess moles of NaOH = 0.024734 mol − 0.0031075 mol = 0.0216265 mol

The total volume of solution after mixing is:

• Total Volume = Volume HCl + Volume NaOH = 28.5 mL + 298.0 mL = 326.5 mL
• Total Volume (in liters) = 0.3265 L

Lastly, we calculate the final concentration of OH− ions:

• Concentration OH− = Moles of excess NaOH / Total Volume (in liters)
• Concentration OH− = 0.0216265 mol / 0.3265 L = 0.0662 M

Therefore, the concentration of excess OH− ions in the final solution is 0.0662 M.