Question

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

Answer 1

The answer is C. 40.0 mL.

Step-by-step explanation:

To solve for the volume of KOH, start by using the formula
`N_(B)`
`V_(B)` =
`N_(A)`
`V_(A)` 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.

`N_(B)` = 0.050 M KOH

`V_(B)` = ?

`N_(A)` = 0.0100 M HNO3

`V_(A)` = 200. mL

Next, use the formula
`N_(B) V_(B) = N_(A)V_(A)`, and in order to find the volume for the base solution, the formula will have to be derived for
`V_(B)`. The formula will now look like
`V_(B)= (N_(A) V_(A) )/(N_(B) )`.

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

Answer 2

I think it’s B .. not sure