If 15.omL of 0.0250 M aqueous H2SO4 is required to neutralize 10.0 mL of aqueous solution of KOH, what is the molarity of the KOH solution?
BMAVA=AMBVB 2(0. 25)(15)=1(10)(MB) 0.75/10=10MB/10 MB=0.075M ANSWERED
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
.25
Hmm... Let me calculate that for you, but first, let me find my calculator. Ah, here it is, my trusty nose! Okay, let's do this mathemagic.
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!
To find the molarity of the KOH solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH).
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.