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A 100 ml sample of HNO3 was diluted to a volume of 10.0 ml. Then 25 ml of that diluted soultion were needed to neutralize 50.0 ml of 0.60 M KOH. What was concentrated of the original nitric acid?
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Start with what you know. You have 50.0 ml of 0.6 M KOH. You can calculate the number of moles of KOH from that. From the balanced equation between KOH and HNO3 you can calculate the moles of HNO3 that neutralizes the KOH. This is the number of moles in 25 mL. BTW, you cant dilute 100 ml to 10 ml. Some mistake in the problem.
Thank you. I had written the problem wrong it was 10.0 mL sample of HNO3 but I get it now.
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