Exam Review #13

a.       1.2 g NaCl x (1 mol NaCl/58.44 g NaCl) = .021 mol NaCl

.021/.450 = .047 M NaCl

b.      .40(.650) =.26 mol

c.       M1V1=M2V2
          .100(.200)= 12.0(V2)
          .00167 L
          1.67 mL of stock solution
           198.33 mL of water will be needed
           -log(.100)=1pH