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?

Question

1535 views around the world

You can reuse this answer
Creative Commons License

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#