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While working in the laboratory, a technician accidentally knocks over a beaker and spills all 200.0 mL of 2.5 M HCl.
Which of the following would neutralize the spill and not have any base left over?

A) 200.0 mL of 2.5 M Ca(OH)2
B) 100.0 mL of 5.0 M KOH
C) 500.0 mL of 1.0 M NaOH
D) 125.0 mL of 2.0 MCa(OH)2
E) 250.0 mL of 2.0 MCa(OH)2

Respuesta :

pstnonsonjoku

Answer:

500.0 mL of 1.0 M NaOH

Explanation:

The number of moles of acid contained in the spill is given by

n=CV

C= concentration of acid= 2.5 M

V= volume of acid = 200mL

n= 2.5 × 200/1000 = 0.5 moles

Hence 0.5 moles of base is required for complete neutralization. Looking at the options, 500.0 mL of 1.0 M NaOH contains 0.5 moles of NaOH.

The reaction is in 1:1 molar ratio.

NaOH(aq) + HCl(aq) ---------> NaCl(aq) + H2O(l)

Hence 0.5 moles of HCl will react with 0.5 moles of NaOH and exactly neutralize the spilt acid.

annyksl

500.0 mL of 1.0 M NaOH would neutralize the spill and leave no base left, as shown in option C.

How to get to this result?

  • First, we must know the number of moles of acid in the poured solution. To do this, we will answer the equation:

[tex]n=C*V[/tex]

[tex]n= 2.5 * (\frac{200}{1000} ) = 0.5 moles[/tex]

With this result, we can say that 0.5 moles of the base will be needed to neutralize all the acid and leave no base left. Thus, we must repeat the equation presented above for all the answer options presented to find out which one has 0.5 moles of base.

  • Option C is the one with this result. And this can be seen in the equation:

[tex]n=C*V\\n= 1*(\frac{500}{1000}) = 0.5 moles[/tex]

More information about acids and bases is at the link:

https://brainly.com/question/7672942

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