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spin [16.1K]
3 years ago
6

A 50.0-ml sample of 0.200 m sodium hydroxide is titrated with 0.200 m nitric acid. calculate the ph in the titration after you a

dd a total of 60.0 ml of 0.200 m hno3. express the ph of the solution to two decimal places.
Chemistry
1 answer:
attashe74 [19]3 years ago
7 0

Hey there!:

Concentration of NaOH = 0.200 M

Concentration of HNO₃= 0.200 M

Total volume =  50.0 mL + 60.0 mL = 110 mL=> 0.11 L

The neutralization reaction between  NaOH and HNO3 :

OH⁻  + H⁺  ---------->  H₂O

So :

n ( H⁺ ) = 60 mL * 0.200 M / 1000 mL  => 0.012 moles of H⁺

n ( OH⁻ ) = 50 mL 0.200 M / 1000 mL => 0.01 moles of OH⁻

Hence OH⁻ is limiting reagent  .

Remaining moles of  H⁺ = 0.012 - 0.01  =>  0.002 moles

Concentration of H⁺  = 0.002 M / 0.11 L

Concentration of H⁺ = 0.01818 moles/L

Therefore:

pH = - log [ H⁺ ]

pH = - log [ 0.01818 ]

pH = 1.74

Hope that helps!

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<u>Given for this question:</u>

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The balanced equation can be obtained by placing coefficients as needed and making sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side

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From the stoichiometry of the balanced equation:

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