Answer:
pH = 8.477
Explanation:
∴ ni   0.3    -      -
 nf    -     0.3    0.3
∴ ni  0.50      -        0.3      Â
 nf  0.5 - X     X      0.3 + X
∴ Ka = 2.0 E-9 = ([H3O+]*[OBr-]) / [HOBr]
⇒ Ka = 2.0 E-9 = ((X/1L)*(0.3 + X)/1 L) / ((0.5 - X)/1L)
⇒ 2.0 E-9 = ( 0.3X + X² ) / (0.5 - X)
⇒ X² + 0.3X - 1 E-9 = 0
⇒ X = 4.333 E-9 M
according Henderson-Hauselbach:
∴ [OBr-] = 0.3 mol/ 1 L + 4.333 E-9 M = 0.300 M
∴ [HOBr] = 0.5 mol / 1 L - 4.33 E-9 M = 0.500 M
∴ pKa = - log Ka = - Log ( 2.0 E-9 ) = 8.6989
⇒ pH = 8.6989 + Log ( 0.300/0.500 )
⇒ pH = 8.477