What volume of 0.365 N lithium hydroxide is needed to completely neutralize 27.68 mL of 0.559 N nitric acid?

Answer 1

#LiOH(aq) + HNO_3(aq) rarr H_2O(l) + LiNO_3(aq)#

Nitric acid moles?

#27.68xx10^(-3)cancelLxx0.559*mol*cancel(L^(-1))# #=# #1.54xx10^(-2)# #mol#.
So we need #(1.54xx10^(-2)*cancel(mol))/(0.365*cancel(mol)*L^-1)# #=# #4.24xx10^(-2)# #L#
Or #4.24xx10^(-2)# #cancelL# #xx# #1000# #mL# #cancel(L^-1)# #=# #??# #mL#
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Answer 2

To find the volume of lithium hydroxide needed to neutralize the nitric acid, we can use the equation:

[ \text{Volume of } \ce{LiOH} = \frac{{\text{Volume of } \ce{HNO3} \times \text{Normality of } \ce{HNO3}}}{{\text{Normality of } \ce{LiOH}}} ]

Plugging in the values:

[ \text{Volume of } \ce{LiOH} = \frac{{27.68 , \text{mL} \times 0.559 , \text{N}}}{{0.365 , \text{N}}} ]

[ \text{Volume of } \ce{LiOH} = \frac{{15.46712}}{{0.365}} ]

[ \text{Volume of } \ce{LiOH} \approx 42.41 , \text{mL} ]

So, approximately 42.41 mL of 0.365 N lithium hydroxide is needed to completely neutralize 27.68 mL of 0.559 N nitric acid.

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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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