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PLEASE HELP. What volume of 0.050 M of KOH neutralizes 200. mL of 0.0100 M HNO3?
a
20.0 ml
b
80.0 mL
40.0 mL
d
30.0 mL

Sagot :

loveaman804
I think it’s B .. not sure
zygon12313

Answer:

The answer is C. 40.0 mL.

Explanation:

To solve for the volume of KOH, start by using the formula [tex]N_{B}[/tex][tex]V_{B}[/tex] = [tex]N_{A}[/tex][tex]V_{A}[/tex] and label the information given in the question. The B in the formula stands for the base solution, and the A in the formula stands for the acid solution.

[tex]N_{B}[/tex] = 0.050 M KOH

[tex]V_{B}[/tex] =   ?    

[tex]N_{A}[/tex] = 0.0100 M HNO3

[tex]V_{A}[/tex] = 200. mL  

Next, use the formula [tex]N_{B} V_{B} = N_{A}V_{A}[/tex], and in order to find the volume for the base solution, the formula will have to be derived for [tex]V_{B}[/tex]. The formula will now look like [tex]V_{B}= \frac{N_{A} V_{A} }{N_{B} }[/tex].

Then, plug in the information given in the question. The equation will look like [tex]V_{B}= \frac{(0.0100 M HNO_{3} {})(200. mL) }{0.050 M KOH}[/tex]. Finally, solve the equation, and the answer will be 40.0 mL.  

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