Answer: PH= 1.4
Explanation:
n(mole)= concentration(c)× volume(v)
n(HClO4) = 0.08 ×0.03
n(HClO4)= 0.0024 mole
n(Ba(OH)2) = 0.28 × 0.05
n(Ba(OH)2) = 0.014
2HClO4 + Ba(OH)2.....>Ba(ClO4)2 + 2H2O.
Hence, the limiting reagent is perchloric acid. According to the reaction:
0.0024 HClO4 req 1/2× 0.0024 of Ba(OH)2.
n(Ba(OH) reacted= 0.0012
Excess n(Ba(OH)2) = 0.014-0.0012
Excess n(Ba(OH)2) = 0.0128
Conc(Ba(OH)2) unreacted = 0.0128/.32 (vol. of mix is 32ml)
= 0.04M = [H+]
PH = -log[0.04] = 1.397
PH= 1.4