An $80mL$ volume of $NaOH$ of $1.00molL^-1$ concentration was stoichiometrically equivalent to a $70mL$ volume of $HCl$ . What is $[HCl]$ ?

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Answer 1 · 2017-12-14T19:24:35

$[HCl]=1.14*mol*L^-1$ , a tad more concentrated than is the base...

We write the stoichiometric equation that represents the reaction between hydrochloric acid, and sodium hydroxide....

$NaOH(aq) + HCl(aq) rarr NaCl(aq) + H_2O(l)$

And we do this to establish the 1:1 stoichiometry....and so then we use the definition of concentration....

$"Concentration"="Moles of solute (mol)"/"Volume of solution (L)"$

And so....

$"moles of NaOH(aq)"=80.0*mLxx10^-3*L*mL^-1xx1.00*mol*L^-1=0.0800*mol$

And there was an equivalent quantity of moles in a $70.0*mL$ volume of $HCl(aq)$ ...

$[HCl(aq)]=(0.0800*mol)/(70.0*mLxx10^-3*L*mL^-1)=1.143*mol*L^-1$

An 80*mL80*mL volume of NaOHNaOH of 1.00*mol*L^-11.00*mol*L^-1 concentration was stoichiometrically equivalent to a 70*mL70*mL volume of HClHCl. What is [HCl][HCl]?