In: Chemistry
1. Calculate [OH?] for 1.7
1) [OH-] = 2 x 1.7 x 10-3 = 3.4 x 10-3 M
2) No.of moles of LiOH = 2.48/24 = 0.1
[OH-] =n x 1000/V = 0.1 x 1000/260
= 0.384M
3) pOH = -log[OH-]
= -log[0.384]
= 0.41
pH = 14 - pOH = 14 - 0.41 = 13.59
4) NO.of moles of HClO3 = w/mw = 0.45/84 = 0.0053
[H+] = 0.0053/2.8 = 0.0018
pH = -log[H+] = -log[0.0018] = 2.74
5) M1V1 = M2V2
5 x 2 = M2 x 0.5 x 103
M2 = 2 x 10-2
[H+] = 2x 10-2
pH = -log[2 x 10-2]
= 1.69
6) [H+] = M1V1 + M2V2/(V1 + V2)
= 2.5 x 10-2 x 60 + 1 x 10-2 x 160/( 60 + 160)
= 1.4 x 10-2
pH = -log[1.4 x 10-2 ]
= 1.85