Question

the inverse log, or -log, as indicated by p, can be reversed. this converts ph back into h₃o⁺¹, poh back into oh⁻¹, and pkₐ back into kₐ. 1. calculate the h₃o⁺¹ when the ph = 6.666 h₃o⁺¹ = 10⁻ᵖᴴ = 10⁻⁶.⁶⁶⁶ = 1 2. calculate the oh⁻¹ when the poh = 8.888 oh⁻¹ = 10⁻ᵖᴼᴴ = 10⁻⁸.⁸⁸⁸ = 2 3. calculate the kₐ when the pkₐ = 7.777 kₐ = 10⁻ᵖᴷₐ = 10⁻⁷.⁷⁷⁷ = 3 a. 1.67 × 10⁻⁸ b. 2.158 × 10⁻⁷ c. 1.29 × 10⁻⁹

Explanation:

Step 1: Calculate \([H_3O^{+1}]\)

We use the formula \([H_3O^{+1}] = 10^{-pH}\). Given \(pH = 6.666\), so we calculate \(10^{-6.666}\).
\(10^{-6.666}=10^{-7 + 0.334}=10^{0.334}\times10^{-7}\). Since \(10^{0.334}\approx2.158\), then \(10^{-6.666}\approx2.158\times 10^{-7}\). So the answer for the first blank is B. \(2.158\times 10^{-7}\)

Step 2: Calculate \([OH^{-1}]\)

We use the formula \([OH^{-1}]=10^{-pOH}\). Given \(pOH = 8.888\), so we calculate \(10^{-8.888}\).
\(10^{-8.888}=10^{-9+ 0.112}=10^{0.112}\times10^{-9}\). Since \(10^{0.112}\approx1.29\), then \(10^{-8.888}\approx1.29\times 10^{-9}\). So the answer for the second blank is C. \(1.29\times 10^{-9}\)

Step 3: Calculate \(K_a\)

We use the formula \(K_a = 10^{-pK_a}\). Given \(pK_a=7.777\), so we calculate \(10^{-7.777}\).
\(10^{-7.777}=10^{-8 + 0.223}=10^{0.223}\times10^{-8}\). Since \(10^{0.223}\approx1.67\), then \(10^{-7.777}\approx1.67\times 10^{-8}\). So the answer for the third blank is A. \(1.67\times 10^{-8}\)

Answer:

1. B. \(2.158\times 10^{-7}\)
2. C. \(1.29\times 10^{-9}\)
3. A. \(1.67\times 10^{-8}\)