Question

A 25.00 mL solution of 0.03190 M Na2CO3 is titrated with 0.02660 M Ba(NO3)2 . Calculate pBa2+ following the addition of the given volumes of Ba(NO3)2 . The ?sp for BaCO3 is 5.0×10−9 .
15.50 mL pBa2+=
?ep Ba2+=
35.50 mL pBa2+

Answer 1

a)

BaCO3 (s) -----------> Ba2+ (aq) + CO32- (aq) :   ,   Ksp = 5.0 x 10^-9

[CO32-] = 25 x 0.03190 - 15.50 x 0.02660 / 25 + 15.5

             = 9.51 x 10^-3 M

Ksp = [Ba2+][CO32-]

5.0 x 10^-9 = [Ba2+] ( 9.51 x 10^-3)

[Ba2+] = 5.26 x 10^-7 M

pBa2+ = 6.279

b)

Ksp = [Ba2+][CO32-]

5.0 x 10^-9 = [Ba2+]^2

[Ba2+] = 7.07 x 10^-5 M

pBa2+ = 4.151

c)

[Ba2+] = 35.50 x 0.02660 - 25 x 0.03190 / 25 + 35.5

             = 2.42 x 10^-3 M

pBa2+ = 2.615