27.80ml of 0.100 M HNO3 is titrated with 27.35ml of an NaOH solution.
What is the molarity of the NaOH ?
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27.80ml of 0.100 M HNO3 is titrated with 27.35ml of an NaOH solution.
What is the molarity of the NaOH ?
We won't do your homework for you.
This kind of problem makes you determine the moles of acid used... then the moles of base it reacted with... then determine molarity from that.
So... how many moles of acid were used? Use the volume and molarity to determine this.
This is a monoprotic acid, meaning it only gives off one H+... that simplifies things
So for the moles of H+... determine the moles of HO-
Then use moles of HO- and volume to determine molarity of base.
Again... we won't do the work for you. Try it out, tell us what you get.
Always write a chemical equation in tasks like this. Start by writing the equation between the reaction of HNO3 and NaOH and their products, then you'll see the mole ratio. From that, you'll be able to tell how many moles of NaOH was present, and the molarity (or concentration).
Use the formulaQuote:
Originally Posted by kickatinalong
M(acid) * V(acid) = M(base) * V(base)
plug in your information and use algebra to solve for the molarity of the base.
It is absolutely fine to show this formula, but also important to note that this only works when you have reactants that react in a 1:1 relationship, as seen in this problem.Quote:
Originally Posted by Raysmommy
Relying strictly on this formula would often throw wrong answers with students on tests or quizzes who didn't understand that stoichiometry needed to be considered. We'd always give a few problems that would deliberately cause the "plug it into m1v1=m2v2" answers to fail without thinking about stoichiometric ratios. Standardized tests also tend to do this.
But all that said... its still a great formula to use when you keep the need for 1:1 in mind.
All this said, the OP'er hasn't still tried to answer the question... fishing for easy answers perhaps?
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