In a titration of 26.0 mL of 0.840 M H2SO4, __________ mL of a 0.200 M KOH solution is required for neutralization. Express your answer to the nearest 1/10 mL. Hint: remember H2SO4 is a diprotic acid
Write the equation and balance it.
moles H2SO4 = M x L.
Convert moles H2SO4 to moles KOH.
Convert moles KOH to molarity. M = moles KOH/L KOH.
To find out how many mL of a 0.200 M KOH solution is required for neutralization with 26.0 mL of 0.840 M H2SO4, we can use the concept of stoichiometry.
The balanced chemical equation for the neutralization between H2SO4 and KOH is as follows:
H2SO4 + 2KOH -> K2SO4 + 2H2O
From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of KOH. Therefore, the mole ratio between H2SO4 and KOH is 1:2.
To determine the number of moles of H2SO4 used, we can use the formula:
moles = concentration (M) × volume (L)
First, we convert the volume of 0.840 M H2SO4 from mL to L:
26.0 mL = 26.0 / 1000 L = 0.026 L
Then, we calculate the number of moles of H2SO4:
moles of H2SO4 = 0.840 M × 0.026 L = 0.02184 mol
Since the mole ratio between H2SO4 and KOH is 1:2, we can determine the number of moles of KOH required as follows:
moles of KOH = 2 × moles of H2SO4 = 2 × 0.02184 mol = 0.04368 mol
Now we can use the number of moles of KOH to find the volume (in mL) of the 0.200 M KOH solution required for neutralization:
volume (mL) = moles / concentration (M)
volume (mL) = 0.04368 mol / 0.200 M = 0.2184 L * 1000 mL/L = 21.84 mL
To express the answer to the nearest 1/10 mL, we round the result to one decimal place:
Approximately 21.8 mL of the 0.200 M KOH solution is required for neutralization.