How many milliliters of .015 M $NaOH$ are needed to neutralize 50.0 mL of 0.010 M $HNO_3$ (aq)? What compounds are formed after the reaction is complete?

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Answer 1 · 2016-03-09T15:41:55

$V_(NaOH)=33mL$

this is a neutralization reaction between a strong acid $HCl$ and a strong base $NaOH$ . The net ionic equation is:

$H^(+)(aq)+OH^(-)(aq)->H_2O(l)$

since the acid and base are both monoprotic, therefore,

$n_(H^+)=n_(OH^-)=>n_(HCl)=n_(NaOH)$

$C_M=n/V=>n=C_MxxV$

$=>(C_MxxV)_(HCl)=(C_MxxV)_(NaOH)$

$=>V_(NaOH)=((C_MxxV)_(HCl))/((C_M)_(NaOH))=(0.010cancel(M)xx50.0mL)/(0.015cancel(M))=33mL$

How many milliliters of .015 M NaOHNaOH are needed to neutralize 50.0 mL of 0.010 M HNO_3HNO_3 (aq)? What compounds are formed after the reaction is complete?