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Consider the following neutralisation reaction:

$$ \ce{2HNO3 + Ba(OH)2 -> Ba(NO3)2 + 2H2O} \tag{1} $$

I would be glad for help in how to solve this problem with $c_1V_1 = c_2 V_2$, because the molar ratio here is not $1:1$.

What volume of a $\pu{0.15M}$ solution of $\ce{Ba(OH)2}$ is required to neutralise $\pu{45 mL}$ of a $\ce{HNO3}$ with concentration of 0.29M solution?

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    $\begingroup$ 1. "45 Millilitres HNO3 solution..." At what concentration? 2. They an only be naturalised if they contain foreign substances. 3. Show work. $\endgroup$
    – DrMoishe Pippik
    Commented Feb 4 at 0:29
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    $\begingroup$ If you keep using C1V1=C2V2 formula, you will never learn how to solve titration problems. $\endgroup$
    – ACR
    Commented Feb 4 at 2:17
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    $\begingroup$ You can use the equation for H+ and OH-. Or you can go via equivalent substance amounts. Note that nobody would use barium hydroxide to titrate HNO3, it is a made up task. $\endgroup$
    – Poutnik
    Commented Feb 4 at 6:32
  • $\begingroup$ This is why NaOH is usually used to titrate acids it is not necessary to multiply the molarity by 2 to get equivalents of hydroxide. $\endgroup$
    – jimchmst
    Commented Feb 4 at 20:27

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