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2. It took 267 mL of 6.88 M NaOH to neutralize 123 mL of an HNO3 solution. What was the concentration of the acid?

User Askar
by
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1 Answer

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(in liters).

Determine the moles of NaOH used:

Moles of NaOH

=

Molarity

×

Volume (L)

Moles of NaOH=Molarity×Volume (L)

Moles of NaOH

=

6.88

mol/L

×

(

267

mL

/

1000

)

Moles of NaOH=6.88mol/L×(267mL/1000)

Use the balanced chemical equation to find the moles of HNO₃:

The balanced equation for the neutralization reaction between NaOH and HNO₃ is:

NaOH

+

HNO₃

NaNO₃

+

H₂O

NaOH+HNO₃→NaNO₃+H₂O

Since the ratio of moles of NaOH to moles of HNO₃ is 1:1, the moles of HNO₃ are also equal to the moles of NaOH.

Determine the concentration of HNO₃:

Concentration of HNO₃

=

Moles of HNO₃

Volume (L)

Concentration of HNO₃=

Volume (L)

Moles of HNO₃

Concentration of HNO₃

=

Moles of NaOH

123

mL

/

1000

Concentration of HNO₃=

123mL/1000

Moles of NaOH

Now, you can plug in the values and calculate the concentration of HNO₃ in simple terms. Remember to perform the calculations using consistent units (e.g., convert mL to L).

User Vivek Mishra
by
8.0k points