Question

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 add a total of 60.0 ml of 0.200 m hno3. express the ph of the solution to two decimal places.

Answer 1

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!

Answer 2

The excess pH of the solution approximated to 2 decimal place is 1.74

From the question we are told that NaOH concentration is = 0.2 M

We are also told from the question, that HNO₃ concentration is the same as NaOH, which is = 0.2 M

We are given volume of both Sodium Hydroxide and Nitric Acid to be 50 mL and 60 mL respectively. This means that the total volume will be the addition of both.

=  50 + 60 = 110 or we convert to Liters 0.11 L

From theory, we know that there exist a neutralization reaction between  NaOH and HNO₃. We then write down this said equation as a path to solving the question.

OH⁻  + H⁺  --->  H₂O

 

next, we have something like this

n( H⁺ ) = 60 × $\frac{0.2}{1000}$  = 0.012 moles of H⁺

n( OH⁻ ) = 50 × $\frac{0.2}{1000}$ => 0.01 moles of OH⁻

from the calculation above, we agree that OH⁻ is limiting reagent.

Proceeding further, we have

Left over moles of  H⁺ = 0.012 - 0.01, and that is  0.002 moles

on division, we get

H⁺  = 0.002 M / 0.11 L

H⁺ = 0.01818 moles/L

Finally, we apply logarithm to get our final answer

pH = - log [ H⁺ ]

pH = - log [0.01818]

pH = 1.74

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