Consider the acid spill. It is already starting to do nasty things to, say, the floor or counter. So you grab the bottle of 10% NaOH and pour some on the spill. All of a sudden, you get a great deal of heat, and you don't have any visual evidence whether your put on too little or too much. But you have added more liquid to the spill, generated more heat, and will get more damage. You have made a bigger mess, and if you added too much, you then have a neutralization problem to deal with.
And if it is something like a strong sulfuric acid solution, adding sodium hydroxide solution will be extremely exothermic, and you could get some really nasty results.
So now approach the spill with a handful of baking soda. You sprinkle it on the spill. It fizzes, and carbon dioxide is given off. That actually, in a very tiny way, moderates the temperature of the neutralization. And you can keep adding baking soda until the fizzing stops, and then perhaps some water to mix everything well. But what you have done is kept the volume to a minimum, added a neutralization agent that has a visible endpoint (no more gas being given off), and you don't suddenly have a huge amount of highly basic solution because you added too much.
And what is also nice about baking soda is that you can toss some with your hand or even with a spoon, and get some distance from the spill. With a liquid, you have to get much closer
i hope this helped..
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
Answer:
7. .........................
Answer:
0.4 M
Explanation:
Molarity is defined as moles of solute, which in your case is sodium hydroxide,
NaOH
, divided by liters of solution.
molarity
=
moles of solute
liters of solution
Notice that the problem provides you with the volume of the solution, but that the volume is expressed in milliliters,
mL
.
Moreover, you don't have the number of moles of sodium hydroxide, you just have the mass in grams. So, your strategy here will be to
determine how many moles of sodium hydroxide you have in that many grams
convert the volume of the solution from milliliters to liters
So, to get the number of moles of solute, use sodium hydroxide's molar mass, which tells you what the mass of one mole of sodium hydroxide is.
7
g
⋅
1 mole NaOH
40.0
g
=
0.175 moles NaOH
The volume of the solution in liters will be
500
mL
⋅
1 L
1000
mL
=
0.5 L
Therefore, the molarity of the solution will be
c
=
n
V
c
=
0.175 moles
0.5 L
=
0.35 M
Rounded to one sig fig, the answer will be
c
=
0.4 M
Explanation:
