How many milliliters of 0.100 M HNO3 are needed to neutralize the following solutions?
A. 45.0 mL of 0.667 M KOH
B. 58.5 mL of 0.0100 M Al(OH)3
Writ the equations and balance them.
HNO3 + KOH ==> KNO3 + H2O
3HNO3 + Al(OH)3 ==> 3H2O + Al(NO3)3
millimoles KOH = mL x M = 45.0*0.667 = 30.015
Convert mmoles KOH to mmoles HNO3 using the coefficients in the balanced equation. That is done as
mmoles HNO3 = mmoles KOH x (1 mole HNO3/1 mole KOH) = 30.015 x (1/1) = 30.015 mmoles HNO3.
Then M = mmoles/mL and solve for mL = mmoles/M = 30.015/0.100 = 300.15 mL. Round to the correct number of sig figures.
The second part is done the same way but the ratio of acid/base is not 1:1 as it is with KOH/HNO3.
300.15 mL
B. 17.55 ml
When Aluminium is added to KOH then solution is?
A. Well, let's do some acid-base clowning! To figure out how many milliliters of 0.100 M HNO3 you need to neutralize 45.0 mL of 0.667 M KOH, first we need to see how many moles of KOH are present. So, 45.0 mL of 0.667 M KOH gives us 0.0450 L x 0.667 mol/L = 0.030015 moles of KOH.
To neutralize this amount of KOH, you'll need an equal number of moles of HNO3. Since the HNO3 is at a concentration of 0.100 M, you can use the good old C1V1 = C2V2 equation.
So, (0.100 M)(V1) = (0.030015 mol)
Solving for V1, you'll get V1 = 0.300 mL. Therefore, you'll need 0.300 mL of 0.100 M HNO3 to neutralize 45.0 mL of 0.667 M KOH.
B. Now, let's move on to the next solution! To neutralize 58.5 mL of 0.0100 M Al(OH)3, you need to find out how many moles of Al(OH)3 are present. So, 58.5 mL of 0.0100 M Al(OH)3 gives you 0.0585 L x 0.0100 mol/L = 0.000585 moles of Al(OH)3.
Since the stoichiometry between Al(OH)3 and HNO3 is 1:3, you'll need three times the number of moles of Al(OH)3 of HNO3 to neutralize it. Therefore, you'll need 0.000585 mol x 3 = 0.001755 moles of HNO3.
And since the HNO3 is at a concentration of 0.100 M, we can use the good old C1V1 = C2V2 equation again.
So, (0.100 M)(V1) = (0.001755 mol)
Solving for V1, you'll get V1 = 0.01755 L. Convert that to milliliters, and you'll need 17.55 mL of 0.100 M HNO3 to neutralize 58.5 mL of 0.0100 M Al(OH)3.
Hope that brings a little acid-base humor into your day!
To find the number of milliliters of 0.100 M HNO3 needed to neutralize a solution, you need to use the concept of stoichiometry.
A. To determine the volume of 0.100 M HNO3 needed to neutralize 45.0 mL of 0.667 M KOH, you need to set up a balanced equation for the neutralization reaction:
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
From the balanced equation, you can see that the ratio of HNO3 to KOH is 1:1. This means that one mole of HNO3 reacts with one mole of KOH. Using this information, you can set up a stoichiometric equation:
(0.100 mol/L HNO3) × (x L HNO3) = (0.667 mol/L KOH) × (45.0 mL KOH)
Solving for x, the volume of 0.100 M HNO3 needed:
x = (0.667 mol/L KOH) × (45.0 mL KOH) / (0.100 mol/L HNO3)
x = 29.985 mL HNO3
Therefore, approximately 30.0 mL of 0.100 M HNO3 are needed to neutralize 45.0 mL of 0.667 M KOH.
B. To determine the volume of 0.100 M HNO3 needed to neutralize 58.5 mL of 0.0100 M Al(OH)3, you need to set up a balanced equation for the neutralization reaction:
3 HNO3(aq) + Al(OH)3(aq) → Al(NO3)3(aq) + 3 H2O(l)
From the balanced equation, you can see that the ratio of HNO3 to Al(OH)3 is 3:1. This means that three moles of HNO3 react with one mole of Al(OH)3. Using this information, you can set up a stoichiometric equation:
(0.100 mol/L HNO3) × (x L HNO3) = (0.0100 mol/L Al(OH)3) × (58.5 mL Al(OH)3) / (1000 mL/L)
Solving for x, the volume of 0.100 M HNO3 needed:
x = (0.0100 mol/L Al(OH)3) × (58.5 mL Al(OH)3) / (1000 mL/L) / (0.100 mol/L HNO3)
x = 0.585 mL HNO3
Therefore, approximately 0.585 mL of 0.100 M HNO3 are needed to neutralize 58.5 mL of 0.0100 M Al(OH)3.