Question

28. A. The acid ionization constant for Cu(H₂O)₆²⁺(aq) is 1.0×10^-7. Calculate the pH of a 0.0349 M solution of Cu(NO₃)₂.

pH = _____

B. The acid ionization constant for Cu(H₂O)₆²⁺(aq) is 1.0×10^-7. Calculate the pH of a 0.0456 M solution of Cu(NO₃)₂.

pH = _____

C. The acid ionization constant for Pb(H₂O)₆²⁺(aq) is 6.3×10^-7. Calculate the pH of a 0.0139 M solution of Pb(NO₃)₂.

pH = _____

Answer 1

A)

Given that ,

C = 0.0349 M

Ka = 1.0x10^-7

pka = -log Ka

= -log(1.0x 10^-7)

= 7.00

Let us consider,

pH = (pKa-logC) /2

= (7- log (0.0349))/2

= (7 - (-1.45)) / 2

= 8.45 / 2

pH = 4.22

B)

Given data , Ka = 1.0x10^-7

C = 0.0456 M

pka = -log Ka

= -log(1.0x 10^-7)

= 7.00

Let us consider ,

pH = (pKa-logC) /2

= (7-log (0.0456))/2

pH = 4.17

C)

Given data,

Ka = 6.3 x10^-7

C = 0.0139 M

pka = -log Ka

= -log(6.3 x10^-7)

= 6.20

Let us consider,

pH = (pKa-logC) /2

= (6.20 - log (0.0624))/2

pH = 3.7