Respuesta :

One way

120 mL of 1.5 M HClO3
50 ml of 1.1 M LiOH

Vf HCLO3 =0.705

Vf LiOH=0.294

PH of solution = -log10 [ 1.5] X 0.705+14 -log10 [1.1] X 0.294

=13.8

Second way

M1v1+m2v2/V1+v2
= 120*1.5+ 50*1.1/ 170
= 1.38M

I hope that helped :)

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