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What volume of 0.150M HClO4
NaOH ?
solution is needed to neutralize 54.00mL of 8.80×10−2M
It helps to write the equation for the reaction:
HClO4 + NaOH ----> NaClO4 + H2O
From this we can tell that it is 1:1 reaction. I like to use a dimensional analysis approach to solving
these. As a general rule, you should start with the compound about which you have the most
information. In this case, it’s NaOH (a volume and concentration).
Use a flowchart like this to figure out what you want…
Quantity A ----> moles A -----> moles B -----> Quantity B
(A is what you star with, B is what you want)
A = NaOH
B = HClO4
54.00 mL A x
8.80 x 10-2 mol A
1LA
x
1 mol B x
1 mol A
1LB
= 31.7 mL B
0.150 mol B
From the flowchart, the first mission is to find moles A (moles A = concentration A x volume A)
The next step is to find moles B. This where the 1:1 mole ratio comes in. Put the coefficients from
the reaction in front of the mol B and mol A.
After that, to get volume of B, we effectively divide by the concentration. This is the same as
multiplying by the “flipped” concentration 0.150 mol/L is flipped to give 1 L/0.150 mol.
Everything cancels to leave us with volume of B.
There are 3 significant figures in 8.80 x 10-2 so there are 3 significant figures in the answer.
What volume of 0.120M HCl is needed to neutralize 2.78g of Mg(OH)2?
Mg(OH)2 + 2 HCl ----> MgCl2 + 2 H2O
Start with Mg(OH)2 because you want HCl.
A = Mg(OH)2
B = HCl
2.78 g A x
1 mol A x
58.32 g A
2 mol B x
1 mol A
1LB
= 0.794 L B
0.120 mol B
The first step was to find moles of A using the molecular weight.
The “2” and “1” in front of the moles comes from the reaction equation.
If 25.0 mL AgNO3 of is needed to precipitate all the Cl- ions in a 0.785-mg sample
of KCl(forming AgCl ), what is the molarity of the AgNO3 solution?
1 KCl + 1 AgNO3 ----> AgCl(s) + KNO3
A = KCl
B = AgNO3
7.85 x 10-3 g A x 1 mol A x
74.55 g A
1 mol B
1 mol A
x1
x
25.0 mL B
1000 mL = 4.21 x 10-4 mol/L B
1L
We need moles/volume, so we divide by volume in the last step to get mol/L.
If 45.5mL of 0.118M HCl solution is needed to neutralize a solution of KOH, how many
grams of KOH must be present in the solution?
A = HCl
B = KOH
HCl + NaOH ----> NaCl + H2O
45.5 mL A x
1L
1000 mL
0.118 mol A
1LA
x
Same basic method as above, except that we need mass.
1 mol B x
1 mol A
56.00 g
1 mol B
= 0.301 g B