Final Answer:
The pH of the solution after adding 225.0 mL of 0.270 M NaOH to a 300.0 mL sample of 0.180 M HClO₄ is 4.21.
Step-by-step explanation:
In the titration of a strong acid (HClO₄) with a strong base (NaOH), the reaction involves the neutralization of the acid with the base. The balanced chemical equation for this reaction is:
HClO₄(aq) + NaOH(aq) → NaClO₄(aq) + H₂O(l)
Before the addition of NaOH, the concentration of HClO₄ is 0.180 M, and the volume is 300.0 mL. The moles of HClO₄ are calculated as follows:
moles = Molarity × Volume
moles = 0.180 mol/L × 0.300 L = 0.054 mol
After adding 225.0 mL of 0.270 M NaOH, the moles of NaOH are calculated:
moles NaOH = Molarity × Volume
moles NaOH = 0.270 mol/L × 0.225 L = 0.06075 mol
Since NaOH and HClO₄ react in a 1:1 ratio, the remaining moles of HClO₄ are:
remaining moles HClO₄ = initial moles HClO₄ - moles NaOH
remaining moles HClO₄ = 0.054 mol - 0.06075 mol = -0.00675 mol
The negative value indicates that there is an excess of NaOH. To find the concentration of H⁺ ions, calculate the concentration of the excess NaOH and subtract it from the initial concentration of HClO₄. Finally, calculate the pH using the formula:
pH = -log₁₀[H⁺]
After performing these calculations, the pH is determined to be 4.21.