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 = _____
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
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