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: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

i think so it’s will be b yes i believe its