i have another one i can't get the write answer for...
Calculate the pH of the colution that results when 27.38 mL of 0.15 M H2SO4 is mixed with 18.50 mL of 0.19 M Ca(OH)2?
Thank you guys so much!!
To calculate the pH of the solution, we need to determine the concentration of hydrogen ions (H+) or hydroxide ions (OH-) in the resulting solution. We can do this by first determining the moles of acid (H2SO4) and base (Ca(OH)2) used.
Step 1: Calculate the moles of H2SO4:
Moles of H2SO4 = Volume (in liters) × Concentration
= 27.38 mL × (1 L / 1000 mL) × 0.15 M
= 0.004107 mol
Step 2: Calculate the moles of Ca(OH)2:
Moles of Ca(OH)2 = Volume (in liters) × Concentration
= 18.50 mL × (1 L / 1000 mL) × 0.19 M
= 0.003515 mol
Step 3: Determine the limiting reactant:
Since H2SO4 and Ca(OH)2 react in a 1:2 ratio (based on their balanced chemical equation), the limiting reactant is Ca(OH)2. This means all of the Ca(OH)2 will react, and some of the H2SO4 will be left.
Step 4: Calculate the excess moles of H2SO4:
Excess moles of H2SO4 = Moles of H2SO4 - (2 × Moles of Ca(OH)2)
= 0.004107 mol - (2 × 0.003515 mol)
= -0.003923 mol
Since we have a negative value for the excess moles of H2SO4, it means that there is no excess H2SO4 remaining after the reaction.
Step 5: Calculate the moles of H+ (hydrogen ions) produced:
Moles of H+ = 2 × Moles of Ca(OH)2
= 2 × 0.003515 mol
= 0.007030 mol
Step 6: Calculate the concentration of H+ ions in the resulting solution:
Concentration of H+ = Moles of H+ / Total Volume of the Solution (in liters)
= 0.007030 mol / ((27.38 mL + 18.50 mL) × (1 L / 1000 mL))
= 0.165 M
Step 7: Calculate the pH:
pH = -log10(concentration of H+)
= -log10(0.165)
≈ 0.782
Therefore, the pH of the resulting solution is approximately 0.782.