Question

2a) The acid ionization constant for Fe(H₂O)₆²⁺(aq) is 3.0×10^-6. Calculate the pH of a 0.0201 M solution of Fe(NO₃)₂.

pH = ______.

2b) The acid ionization constant for Cu(H₂O)₆²⁺(aq) is 1.0×10^-7. Calculate the pH of a 0.0249 M solution of Cu(NO₃)₂.

pH = _____.

2c) The acid ionization constant for Cu(H₂O)₆²⁺(aq) is 1.0×10^-7. Calculate the pH of a 0.0356 M solution of Cu(NO₃)₂.

pH = _____.

Answer 1

[H⁺] = (K_a[C])^1/2

Where, [H⁺] is the concentration of H⁺

K_a is the acid dissociation constant

And C is the concentration of complex

SInce in each complex, M(H₂O)₆⁺ has 1 mole of Metal. So, the concentration of complex = concentration of metal.

2a) K_a = 3.0 *10^-6

C = 0.0201 M

So, [H⁺] = ( 3.0 *10^-6 x 0.0201 M)^1/2 = 2.46*10^-4

pH = -log[H⁺] = -log (2.46*10^-4) = 3.61

2b)K_a = 1.0 *10^-7

C = 0.0249 M

So, [H⁺] = ( 1.0 *10^-7 x 0.0249 M)^1/2 = 4.99*10^-5

pH = -log[H⁺] = -log (4.99*10^-5) = 4.30

2c) K_a = 1.0 *10^-7

C = 0.0356 M

So, [H⁺] = ( 1.0 *10^-7 x 0.0356 M)^1/2 = 5.97*10^-5

pH = -log[H⁺] = -log (5.97*10^-5) = 4.22