A solution of KOH is prepared by dissolving 2.00 g of KOH in water to a final volume of 250 ml of solution what volume of this solution will neutralize 20.0 ml of 0.115 mol/L sulfuric acid?
amount of KOH= 2.00g(1mol of KOH/ 56.01g)= 0.357 moles
concentration of KOH in water = amount of KOH/ volume of kOH= ( 0.357 mol/(25x10^-3)= 14 mol/ l
please help
molar mass KOH is 56.1 for starters and
2.00/56.1 = 0.0357
M KOH = 0.0357/0.250 = 0.143 instead of your number of 14.
H2SO4 + 2KOH ==> K2SO4 + + 2H2O
mols H2SO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols H2SO4 to mols KOH.
Then M KOH = mols KOH/L KOH. YOu know M KOH and mols KOH, solve for L KOH
To determine the volume of the KOH solution needed to neutralize the sulfuric acid, we can use the equation:
Molarity (mol/L) x Volume (L) = Molarity (mol/L) x Volume (L)
First, calculate the number of moles of sulfuric acid:
Moles of sulfuric acid = Molarity x Volume
= 0.115 mol/L x 0.020 L
= 0.0023 mol
Since KOH reacts with sulfuric acid in a 1:2 ratio, the number of moles of KOH needed will be twice the moles of sulfuric acid:
Moles of KOH = 2 x Moles of sulfuric acid
= 2 x 0.0023 mol
= 0.0046 mol
Now, let's calculate the volume of the KOH solution needed:
Volume (L) = Moles of KOH / Molarity
= 0.0046 mol / 14 mol/L (concentration of KOH in water)
≈ 0.0003286 L
To convert this volume to milliliters (ml):
Volume (ml) = 0.0003286 L x 1000 ml/L
≈ 0.3286 ml
Therefore, approximately 0.3286 ml of the KOH solution is needed to neutralize 20.0 ml of 0.115 mol/L sulfuric acid.
To find the volume of the KOH solution required to neutralize the given amount of sulfuric acid, we can use the concept of stoichiometry.
1. Write the balanced chemical equation for the reaction between KOH and sulfuric acid:
2 KOH + H2SO4 -> K2SO4 + 2 H2O
2. Use the balanced equation to determine the mole ratio between KOH and sulfuric acid. From the equation, we can see that 2 moles of KOH react with 1 mole of sulfuric acid.
3. Calculate the number of moles of sulfuric acid in the given 20.0 ml of 0.115 mol/L sulfuric acid:
Moles of sulfuric acid = concentration x volume
Moles of sulfuric acid = 0.115 mol/L x 0.020 L = 0.0023 moles
4. Using the mole ratio from step 2, we can determine the number of moles of KOH needed to neutralize the sulfuric acid. Since 2 moles of KOH react with 1 mole of sulfuric acid, we need half the number of moles of KOH:
Moles of KOH needed = 0.0023 moles / 2 = 0.00115 moles
5. Now, we can use the concentration of the KOH solution to convert the moles of KOH needed into volume:
Volume of KOH solution = Moles of KOH needed / Concentration of KOH solution
Plugging in the values:
Volume of KOH solution = 0.00115 moles / (14 mol/L) = 8.21 x 10^-5 L
6. Finally, convert the volume from liters to milliliters:
Volume of KOH solution = 8.21 x 10^-5 L x 1000 = 0.0821 mL
Therefore, approximately 0.0821 mL of the KOH solution is required to neutralize 20.0 mL of 0.115 mol/L sulfuric acid.