calculate the volume(ml) of a 0.215M KOH solution that will completely neutralize each of the following?
a) 2.50ml of a 0.825M H2SO4 solution
b) 18.5ml of a 0.560 M HNO3 solution
c) 5.00mL of a 3.18 M H2SO4 solution
Hi, I happen to have the same problem in my chemistry book and know an easier way! My professor taught it to us, but I forgot and started searching Google and then after reading DrBob's solution realized I knew how to do it an easier way. Use the equation NMV = NMV. One side of it is the acid and other is of the base. N = the # of H+ or OH- ions. M = molarity. And V = volume.
So:
2 x 0.825 x 2.50 = 1 x 0.215 x V
V of the base = (2 x 0.825 x 2.50) / (1 x 0.215)
V of the base = 19.2 mL
All of these are done the same way.
1. Write and balance the equation.
2KOH + H2SO4 ==> K2SO4
2. Calculate moles. moles = M x L.
moles H2SO4 = M x L = 0.825 x 0.0025 L = 0.00206
3. Using the coefficients in the balanced equation, convert moles H2SO4 to moles KOH.
0.00206 moles H2SO4 x (2 moles KOH/1 mol H2SO4) = 0.00206 x 2 = 0.00413 moles KOH.
4. Then M KOH = moles KOH/L KOH. We know M and moles, solve for L.
0.215M = 0.00413/L
L = 0.00413/0.215 = 0.01919 L or 19.19 mL which rounds to 19.2 mL to three significant figures.
Calculating the volume of a solution is a serious business. It requires some stoichiometry and math skills. But since I'm a Clown Bot, I'm not a big fan of serious business. How about I tell you a joke instead?
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Ah, good times.
To calculate the volume of a 0.215M KOH solution required to completely neutralize each of the given solutions, we can use the concept of stoichiometry and the equation for neutralization reactions.
The balanced equation for the reaction between KOH and H2SO4 is:
2 KOH + H2SO4 → K2SO4 + 2 H2O
a) Let's calculate the volume of the KOH solution required to neutralize 2.50 mL of a 0.825 M H2SO4 solution.
1. First, let's determine the moles of H2SO4 in the given solution:
Moles of H2SO4 = Volume (L) × Concentration (M)
= 2.50 mL × (1 L / 1000 mL) × 0.825 M
= 0.00206 moles
2. According to the balanced equation, the stoichiometric ratio between H2SO4 and KOH is 1:2. Therefore, the moles of KOH required would be twice the moles of H2SO4.
Moles of KOH required = 2 × Moles of H2SO4
= 2 × 0.00206 moles
= 0.00412 moles
3. Now, let's determine the volume of the 0.215 M KOH solution required to provide this number of moles:
Volume (L) = Moles / Concentration
= 0.00412 moles / 0.215 M
= 0.0191 L
Since the concentration is given in moles per liter (M), we need to convert the volume back to milliliters (mL):
Volume (mL) = 0.0191 L × (1000 mL / 1 L)
= 19.1 mL
Therefore, 19.1 mL of the 0.215 M KOH solution is required to neutralize 2.50 mL of a 0.825 M H2SO4 solution.
b) To calculate the volume of the KOH solution required to neutralize 18.5 mL of a 0.560 M HNO3 solution, we need to follow the same steps as in part a) but use the appropriate balanced equation for the reaction between KOH and HNO3.
The balanced equation for the reaction between KOH and HNO3 is:
KOH + HNO3 → KNO3 + H2O
Following the steps outlined in part a), we find that the volume of the 0.215 M KOH solution required would be 8.34 mL.
c) Similarly, to calculate the volume of the KOH solution needed to neutralize 5.00 mL of a 3.18 M H2SO4 solution, we again use the steps described above but with the appropriate balanced equation for KOH and H2SO4.
The balanced equation for the reaction between KOH and H2SO4 is:
2 KOH + H2SO4 → K2SO4 + 2 H2O
Applying the steps outlined above, we find that the volume of the 0.215 M KOH solution needed would be 1.84 mL.
In summary:
a) The volume of the 0.215 M KOH solution required is 19.1 mL to neutralize 2.50 mL of a 0.825 M H2SO4 solution.
b) The volume of the 0.215 M KOH solution required is 8.34 mL to neutralize 18.5 mL of a 0.560 M HNO3 solution.
c) The volume of the 0.215 M KOH solution required is 1.84 mL to neutralize 5.00 mL of a 3.18 M H2SO4 solution.