Final answer:
To calculate the pH and pOH of a mixture of sodium hydroxide and sulphuric acid, moles of each reactant are determined, and the excess of H+ or OH- ions is calculated. The pH and pOH of the mixture are found using the concentration of the excess ions.
Step-by-step explanation:
To find the pH and pOH of a mixture containing 15.0ml of 0.126M sodium hydroxide and 21.0ml of 0.051M sulphuric acid, we need to first understand the reaction between the acid and the base. Sodium hydroxide (NaOH) is a strong base, and sulphuric acid (H2SO4) is a strong acid. Sulphuric acid is also diprotic, meaning it can donate two protons (H+) per molecule.
First, we need to calculate the moles of each reactant:
Moles of NaOH: 15.0 mL × 0.126 M = 0.00189 molesMoles of H2SO4: 21.0 mL × 0.051 M = 0.001071 moles
Since sulphuric acid can donate two protons, we multiply the moles of H2SO4 by 2 to get the total moles of H+ ions:
Total moles of H+ from H2SO4: 0.001071 moles × 2 = 0.002142 moles
We then subtract the moles of H+ from the moles of OH− (× 1, as sodium hydroxide releases one hydroxide ion per molecule) to see which one has an excess:
Excess moles of OH−: 0.00189 moles - 0.002142 moles = -0.000252 moles
Since we got a negative value, it means we have excess H+ ions, equal to 0.000252 moles.
To find the final concentration of H+, we need to add up the volumes of the solutions (remembering that 1 mL = 1 cm3):
Total volume: 15.0 mL + 21.0 mL = 36.0 mL = 0.036 L
[H+]: = 0.000252 moles / 0.036 L = 0.007 M
Now we can find the pH and pOH:
pH: = -log(0.007) = 2.15pOH: = 14 - pH = 14 - 2.15 = 11.85
The pH and pOH of the reaction mixture after neutralization are approximately 2.15 and 11.85, respectively.