What is the concentration of a #KOH (aq)# solution if 12.8 mL of this solution is required to react with 25.0 mL of .110 mol/L #H_2SO_4(aq)#?

Answer 1

We assume stoichiometric quantities of potassium hydroxide and sulfuric acid.

#2KOH(aq) + H_2SO_4 rarr K_2SO_4(aq) + 2H_2O(l)#

The reaction is an example of potassium hydroxide neutralizing sulfuric acid.

Moles of sulfuric acid: #25.0xx10^-3cancelLxx0.110*mol*cancel(L^-1)# #=# #??# #mol#
From the equation above, which shows that the stoichiometry of hydroxide to sulfuric acid is #2:1#, and the volume of stoichiometric acid:
#[KOH]=(25.0xx10^-3cancelLxx0.110*mol*cancel(L^-1)xx2)/(12.8xx10^-3L)# #=# #??mol*L^-1#.
We get an answer in #mol*L^-1#, as required.
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Answer 2

The concentration of the KOH (aq) solution is 0.055 mol/L.

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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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