A50.0-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.

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; economic order quantity;

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole