how many mL of a 0.20M KOH solution is needed to neutralize 25 mL of 0.25 M H2SO4?
H2SO4 + 2NaOH ==>Na2SO4 + 2H2O
mols H2SO4 = M x L = ??
Look at the coefficients in the balanced equation. mols NaOH = 2 x mole H2SO4.
Then M NaOH = moles NaOH/L NaOH
You know M and you know moles, solve for L NaOH and convert to mL.
To solve this problem, we can use the concept of stoichiometry, which involves the balanced chemical equation and the ratios of the reacting species. In the given question, we need to determine the volume of a 0.20 M KOH solution required to neutralize 25 mL of 0.25 M H2SO4.
Step 1: Write the balanced chemical equation for the neutralization reaction between KOH and H2SO4:
2 KOH + H2SO4 -> K2SO4 + 2 H2O
Step 2: Determine the mole to mole ratio from the balanced equation:
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of H2SO4.
Step 3: Calculate the number of moles of H2SO4:
Moles of H2SO4 = Concentration of H2SO4 (in M) × Volume of H2SO4 (in L)
Moles of H2SO4 = 0.25 M × 0.025 L = 0.00625 moles
Step 4: Calculate the number of moles of KOH required:
From the mole to mole ratio, we know that 2 moles of KOH react with 1 mole of H2SO4. So to neutralize 0.00625 moles of H2SO4, we would need (2/1) × 0.00625 moles of KOH.
Moles of KOH = (2/1) × 0.00625 moles = 0.0125 moles
Step 5: Calculate the volume of the 0.20 M KOH solution needed:
Moles of KOH = Concentration of KOH (in M) × Volume of KOH (in L)
0.0125 moles = 0.20 M × Volume of KOH (in L)
Volume of KOH (in L) = 0.0125 moles / 0.20 M = 0.0625 L
Since we want the volume in mL, we multiply the volume in L by 1000:
Volume of KOH (in mL) = 0.0625 L × 1000 = 62.5 mL
Therefore, 62.5 mL of the 0.20 M KOH solution is needed to neutralize 25 mL of the 0.25 M H2SO4 solution.