A chemist dissolves 751.mg of pure nitric acid in enough water to make up 290.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.

1.4

Explanation:

Mass of pure nitric acid = 751mg

Volume of solution  = 290mL

Unknown:

pH of the solution  = ?

Solution:

To solve this problem, we need the concentration of the acid in the aqueous form.

   This is given by molarity;

               Molarity  =

Since the number of moles of nitric acid is unknown, we can easily solve for it.

      Number of moles of nitric acid  =

            molar mass of HNO₃   =  1 + 14 + 3(16)  = 63g/mol

             mass of nitric acid  = 751mg  = 0.751g

     Number of moles  =    = 0.012mole

Volume of solution = 290mL  = 0.29dm³

Now molarity of the solution  =   = 0.041moldm⁻³

Since:

     pH  = -log [H₃O⁺]

         HNO₃   +     H₂O  →    H₃O⁺         +         NO₃⁻

      1moldm⁻³                     1moldm⁻³            1moldm⁻³

   0.041moldm⁻³             0.041moldm⁻³    0.041moldm⁻³

  pH  = -log[0.041]   = 1.4

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g