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?

Answer 1

#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)#

Because both the acid and the base are monoprotic,

#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#
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Answer 2

To neutralize 50.0 mL of 0.010 M HNO3, you need 0.025 moles of NaOH. Therefore, you need 1.67 mL of 0.015 M NaOH. The compounds formed after the reaction are NaNO3 and H2O.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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