Question

A 1.00 g piece of Mn metal (54.94 g/mol) was placed in 200. mL of a nitric acid solution with an initial pH of 0.602. After all of the Mn metal has reacted (producing hydrogen gas and Mn2+ and NO3- ions), what should be the final pH of the solution? (Assume constant volume.) 1.17 0.80 1.87 0.74

Answer 1

Sol . As Reaction :

Mn + 2HNO3 ---> Mn2+ + 2NO3- + H2

Now , initial pH = 0.602

So , initial concentration of HNO3 = [HNO3] =[H+] = antilog(-initial pH) = antilog(-0.602) = 0.250 M

As volume of solution = 200 mL = 0.2 L

So ,initial moles of HNO3 = Conc. × volume = 0.250 × 0.2 = 0.05 moles

Now , mass of Mn = 1.00 g

molar mass of Mn = 54.94 g/mol

So, moles of Mn = mass/molar mass = 1.00/54.94 = 0.0182 moles

Now , from the reaction , 1 mole of Mn combines with 2 moles of HNO3 .

So, 0.0182 moles of Mn combines with  = 2 × 0.0182 = 0.0364 moles of HNO3 .

Therefore , moles of HNO3 reacted = 0.0364 moles

and , moles of HNO3 unreacted = initial moles of HNO3 - moles of HNO3 reacted = 0.05 - 0.0364 = 0.0136 moles

Now , volume of solution = 0.2 L

So, final conc. of HNO3 = [H+] = moles/volume

=  0.0136 /  0.2 = 0.068 M

Therefore , final pH = -log[H+] = - log(0.068) = 1.17

So, Answer is 1.17