A 20.0 mL sample of 0.432 M HBr is titrated with a 0.314 M NaOH solution....

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

Solutions

Expert Solution

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

queen_honey_blossom answered 1 month ago

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