In: Chemistry
A 20.0 mL sample of 0.432 M HBr is titrated with a 0.314 M NaOH solution. The pH after the addition of 35.0 mL of NaOH is __________.
A.0.04 | |
B.1.37 | |
C.12.53 | |
D.12.63 | |
E.13.33 |
answer : option D) 12.63
millimiles of acid (HBr) = 20 x 0.432
= 8.64
millimoles of base (NaOH) = 35 x 0.314
= 10.99
millimoles of base > milli moles of acid
here base millimoles dominates the acid millimoles so
[OH-] = base millimoles - acid millimoles / total volume
= 10.99 - 8.64 / (20 + 35)
= 0.0427 M
pOH = -log [OH-]
= -log (0.0427)
= 1.37
pH + pOH = 14
pH = 14 - pOH
= 14 - 1.37
= 12.63
pH = 12.63