Question

PLEASE HELP. What volume of 0.050 M of KOH neutralizes 200. mL of 0.0100 M HNO3?

Answer 1

I think it’s B .. not sure

Answer 2

Answer:

The answer is C. 40.0 mL.

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 $V_{B}= \frac{N_{A} V_{A} }{N_{B} }$.

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.