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:(
Yes, both are wrong. And you posted that same question yesterday and I answered it telling you how to do the a part. The b part is done the same way.
I keep getting the same thing for b. .17
To determine the volume of KOH solution required to neutralize the given samples of sulfuric acid, we need to use the balanced neutralization reaction and the concept of stoichiometry.
Let's start by calculating the number of moles of sulfuric acid present in each sample using the given molarity (M) and volume (mL).
a) 26 mL of 0.240 M H2SO4:
To find the number of moles, we need to convert the volume from mL to L.
Volume of H2SO4 = 26 mL = 26/1000 L = 0.026 L
Number of moles of H2SO4 = Molarity x Volume
Number of moles of H2SO4 = 0.240 M x 0.026 L = 0.00624 moles
b) 143 mL of 0.14 M H2SO4:
Again, convert the volume from mL to L.
Volume of H2SO4 = 143 mL = 143/1000 L = 0.143 L
Number of moles of H2SO4 = Molarity x Volume
Number of moles of H2SO4 = 0.14 M x 0.143 L = 0.02002 moles
Now that we know the number of moles of H2SO4 in each sample, we can use the stoichiometry of the balanced reaction to determine the moles of KOH required.
According to the balanced reaction:
1 mole of H2SO4 reacts with 2 moles of KOH.
a) For 0.00624 moles of H2SO4, the molar ratio with KOH is 1:2.
Therefore, the moles of KOH required = 0.00624 moles x 2 = 0.01248 moles.
b) For 0.02002 moles of H2SO4, the molar ratio with KOH is 1:2.
Therefore, the moles of KOH required = 0.02002 moles x 2 = 0.04004 moles.
Now, to find the volume of the 0.240 M KOH solution required, we need to calculate the volume using the moles of KOH and the given molarity.
Volume of KOH = Moles of KOH / Molarity of KOH
a) For 0.01248 moles of KOH:
Volume of KOH = 0.01248 moles / 0.240 M = 0.052 L = 52 mL (rounded to three significant figures)
b) For 0.04004 moles of KOH:
Volume of KOH = 0.04004 moles / 0.240 M = 0.166 L = 166 mL (rounded to three significant figures)
So, the correct answers are:
a) The volume of 0.240 M KOH solution required to neutralize 26 mL of 0.240 M H2SO4 is 52 mL.
b) The volume of 0.240 M KOH solution required to neutralize 143 mL of 0.14 M H2SO4 is 166 mL.