2KOH + H2SO4 ==> K2SO4 + 2H2O
moles H2SO4 = M x L = ??
Convert moles H2SO4 to moles KOH. It will be ??mole H2SO4 x 2 = moles KOH.
M KOH = moles KOH/L KOH
moles H2SO4 = M x L = ??
Convert moles H2SO4 to moles KOH. It will be ??mole H2SO4 x 2 = moles KOH.
M KOH = moles KOH/L KOH
To neutralize KOH with H2SO4, we need equal amounts of moles of both. So, we can write the equation like this:
H2SO4 + 2KOH → K2SO4 + 2H2O
The molar ratio is 1:2. Now, let's find the moles of H2SO4 that were initially used:
moles of H2SO4 = Molarity × Volume
= 0.0250 M × 0.0150 L
= 0.000375 moles
Since the molar ratio is 1:2, the moles of KOH will also be equal to 0.000375 moles. We can then calculate the molarity of the KOH solution like this:
Molarity of KOH = moles ÷ Volume
= 0.000375 moles ÷ 0.0100 L
= 0.0375 M
So, the molarity of the KOH solution is 0.0375 M. Voila!
The balanced equation for the reaction is as follows:
H2SO4 + 2 KOH → K2SO4 + 2 H2O
From the equation, we can see that 1 mole of H2SO4 reacts with 2 moles of KOH.
First, let's determine the number of moles of H2SO4 used in the neutralization reaction:
Moles of H2SO4 = volume (L) × molarity (mol/L) = 15.0 mL × 0.0250 mol/L = 0.3750 mmol
Since 1 mole of H2SO4 reacts with 2 moles of KOH, the number of moles of KOH used in the reaction is:
Moles of KOH = 0.3750 mmol ÷ 2 = 0.1875 mmol
Now, let's calculate the molarity of the KOH solution:
Molarity of KOH = moles of KOH ÷ volume (L) = 0.1875 mmol ÷ 0.0100 L = 18.75 mmol/L
Converting mmol/L to mol/L:
Molarity of KOH = 18.75 mmol/L × (1 mol/1000 mmol) = 0.01875 M
Therefore, the molarity of the KOH solution is 0.01875 M.