Question 15.3: What is the pH of a solution with an [H^+] of (a) 1.0 × 10^-...

(a) [H^+] = 1.0 \times  10^{-11}

(2 significant figures)

pH = -log(1.0 \times  10^{-11})

pH= 11.00

(2 decimal places)

(b) [H^+]= 6.0 \times  10^{-4}

(2 significant figures)

log (6.0\times  10^{-4}) = -3.22

pH= -log[H^+]

pH = -(-3.22) = 3.22

(2 decimal places)

(c) [H^+] = 5.47 \times  10^{-8}

(3 significant figures)

log (5.47 \times  10^{-8}) = -7.262

pH= -log[H^+]

pH = -(-7.262) = 7.262

(3 decimal places)