Question

What volume of a 0.295 M nitric acid solution is required to neutralize 24.8 mL of a 0.107 M potassium hydroxide solution

Answer 1

Answer: 9.00 mL

Step-by-step explanation:

`M_(A)V_(A)=M_(B)V_(B)\\(0.295)V_(A)=(0.107)(24.8)\\V_(A)=((0.107)(24.8))/((0.295)) \approx \boxed{9.00 \text{ mL}}`