Neutralization occurs when $15.0 m L$ $15.0mL$ of $K O H$ $KOH$ react with $25.0 m L$ $25.0 mL$ of $H N O 3$ $HNO3$ . If the molarity of $H N O 3$ $HNO3$ is $0.750 M$ $0.750 M$ , what is the molarity of the $K O H$ $KOH$ ?

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Answer 1 · 2018-06-03T20:50:40

$[KOH]=1.25*mol*L^-1....$

We interrogate the reaction....

$KOH(aq) + HNO_3(aq) rarr KNO_3(aq) + H_2O(l)$

$"Moles of nitric acid"=25.0*mLxx10^-3*L*mL^-1xx0.750*mol*L^-1=0.01875*mol$ ...

And an EQUIVALENT molar quantity of $"potassium hydroxide"$ was present in the $15.0*mL$ volume...and thus a concentration of...

$(0.01875*mol)/(15.0*mLxx10^-3*L*mL^-1)=??*mol*L^-1$

Neutralization occurs when 15.0mL15.0mL15.0mL of KOHKOHKOH react with 25.0 mL25.0mL25.0 mL of HNO3HNO3HNO3. If the molarity of HNO3HNO3HNO3 is 0.750 M0.750M0.750 M, what is the molarity of the KOHKOHKOH?