How many Ml of 0.200 M NaOH are needed to neutralize 15.00 mL of the 0.100 M hypothetical acid H2A?
H2A + 2NaOH ==> NaA + 2H2O
mols H2A = M x L = ?
1 mol H2A = 2 mols NaOH; therefore, mols NaOH = mols H2A x 2 = ?
Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and M NaOH, solve for L NaOH and convert to mL.
To determine the number of mL of 0.200 M NaOH needed to neutralize 15.00 mL of the 0.100 M hypothetical acid H2A, you need to use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation for the neutralization reaction between the acid H2A and NaOH is:
H2A + 2NaOH -> Na2A + 2H2O
From the balanced equation, you can see that one molecule of H2A reacts with two molecules of NaOH.
To find the number of moles of H2A in 15.00 mL of the 0.100 M solution, you can use the formula:
moles = Molarity x Volume (in liters)
moles of H2A = 0.100 M x (15.00 mL / 1000 mL/ L) = 0.0015 moles
According to the stoichiometry of the balanced equation, 1 mole of H2A reacts with 2 moles of NaOH. Therefore, the number of moles of NaOH needed can be calculated as follows:
moles of NaOH = 0.0015 moles of H2A x (2 moles of NaOH / 1 mole of H2A) = 0.003 moles
Finally, to determine the volume of 0.200 M NaOH needed to neutralize the acid, you can rearrange the formula for moles:
moles = Molarity x Volume (in liters)
0.003 moles = 0.200 M x Volume (in liters)
Volume (in liters) = 0.003 moles / 0.200 M = 0.015 L
Since 1 L is equivalent to 1000 mL, the volume in mL is:
Volume (in mL) = 0.015 L x (1000 mL / 1 L) = 15 mL
Therefore, you would need 15 mL of 0.200 M NaOH to neutralize 15.00 mL of the 0.100 M hypothetical acid H2A.