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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 :

MEDUNNO13GO

Answer: 9.00 mL

Explanation:

[tex]M_{A}V_{A}=M_{B}V_{B}\\(0.295)V_{A}=(0.107)(24.8)\\V_{A}=\frac{(0.107)(24.8)}{(0.295)} \approx \boxed{9.00 \text{ mL}}[/tex]

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