E sure to answer all parts. Calculate the pH of the following aqueous solutions at 25°C: (a) 9.5 × 10−8 M NaOH (b) 6.3 × 10−2 M LiOH (c) 6.3 × 10−2 M Ba(OH)2

(a) 7.0

(b) 12.8

(c) 13.11

Explanation:

(a)

NaOH is a strong base with 1 OH⁻. If [NaOH] = 9.5 × 10⁻⁸ M, then [OH⁻] = 9.5 × 10⁻⁸ M.

The pOH is:

pOH = -log [OH⁻] = -log 9.5 × 10⁻⁸ = 7.0

The pH is:

pH = 14 - pOH = 14 - 7.0 = 7.0

(b)

LiOH is a strong base with 1 OH⁻. If [LiOH] = 6.3 × 10⁻² M, then [OH⁻] = 6.3 × 10⁻² M.

The pOH is:

pOH = -log [OH⁻] = -log 6.3 × 10⁻² = 1.2

The pH is:

pH = 14 - pOH = 14 - 1.2 = 12.8

(c)

Ba(OH)₂ is a strong base with 2 OH⁻. If [Ba(OH)₂] = 6.3 × 10⁻² M, then [OH⁻] = 2 × 6.3 × 10⁻² M = 0.13 M

The pOH is:

pOH = -log [OH⁻] = -log 0.13 = 0.89

The pH is:

pH = 14 - pOH = 14 - 0.89 = 13.11

m=9

step-by-step explanation:

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2