Question

In: Chemistry

1. Calculate [OH?] for 1.7

Expert Solution

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

queen_honey_blossom answered 4 months ago

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