Determine the volume of 0.240 M KOH solution required to neutralize each of the following samples of sulfuric acid. The neutralization reaction is shown below.
H2SO4(aq) + 2 KOH(aq)---> K2SO4(aq) + 2 H2O(l)
a) 26 mL of 0.240 M H2SO4
b) 143 mL of 0.14 M H2SO4
for a I got .012 and for b I got .17 but they are both wrong:(
I don't know how you came up with those numbers.
2KOH + H2SO4 ==> K2SO4 + 2H2O
moles H2SO4 = M x L = ??
Use the coefficients to change moles H2SO4 to mole KOH. You can see that moles NaOH = 2x moles H2SO4
Then MNaOH = moles NaOH/L NaOH
Then convert L NaOH to mL.
You should obtain 52 mL.
Im confused, where did you get NaOH?
To determine the volume of 0.240 M KOH solution required to neutralize each sample of sulfuric acid, we need to use the stoichiometry of the balanced chemical equation provided.
Let's solve the problem step by step.
a) 26 mL of 0.240 M H2SO4:
First, we need to find the number of moles of H2SO4 in the given sample.
moles of H2SO4 = volume (in L) × concentration (in M)
moles of H2SO4 = 26 mL × (1 L / 1000 mL) × 0.240 M
moles of H2SO4 = 0.00624 mol
Now, according to the balanced chemical equation, the mole ratio between H2SO4 and KOH is 1:2. This means that 1 mole of H2SO4 will react with 2 moles of KOH.
Since the concentration of KOH is also given as 0.240 M, we can use the mole ratio to find the volume of KOH solution required.
moles of KOH = (moles of H2SO4) × (2 moles of KOH / 1 mole of H2SO4)
moles of KOH = 0.00624 mol × (2/1)
moles of KOH = 0.01248 mol
Finally, we can find the volume of KOH solution using the concentration of KOH.
volume of KOH solution = (moles of KOH) / (concentration of KOH)
volume of KOH solution = 0.01248 mol / 0.240 M
volume of KOH solution = 0.052 L or 52 mL
Therefore, the volume of 0.240 M KOH solution required to neutralize 26 mL of 0.240 M H2SO4 is 52 mL.
b) 143 mL of 0.14 M H2SO4:
Following the same steps as above, let's calculate the volume of KOH solution required.
moles of H2SO4 = 143 mL × (1 L / 1000 mL) × 0.14 M
moles of H2SO4 = 0.02002 mol
moles of KOH = 0.02002 mol × (2/1)
moles of KOH = 0.04004 mol
volume of KOH solution = 0.04004 mol / 0.240 M
volume of KOH solution = 0.167 L or 167 mL
Therefore, the volume of 0.240 M KOH solution required to neutralize 143 mL of 0.14 M H2SO4 is 167 mL.
It seems your answers were close, but might have been rounded incorrectly. Make sure to carry extra decimal places throughout the calculation, and round the final answer to the appropriate number of significant figures.