A buffer solution contains 0.368 M hydrocyanic acid and 0.360 M potassium cyanide . If 0.0513 moles of sodium hydroxide are added to 225 mL of this buffer, what is the pH of the resulting solution

Solution :

Millimoles of hydrocyanic acid = 225 x 0.368

                                                   = 82.8

Millimoles of potassium cyanide = 225 x 0.360

                                                   = 81

Millimoles of sodium hydroxide = 51.3

Therefore,

pOH = pKb + log [salt - C / bas + C]

        = 4.74 + log[82.8 - 51.3 / 81 + 51.3]

         = 4.102

Therefore, pH = 9.05

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.