1. Calculate the concentrations of OH - and the pH of each basic solution. Since

Question

1. Calculate the concentrations of OH- and the pH of each basic solution. Since NaOH donates one OH- ion for each NaOH molecule, the OH- concentration is equivalent to the NaOH molarity. The pOH is calculated as -log[OH-] and the pH is calculated as 14 - pOH. Summarize your results for each pH indicator, recording both the pH values and the color of the solution at each pH.

How do you figure the OH-, pOH, and the pH?

Tube 1: (.1M Sodium Hydroxide)(5mL) = (?)(10mL)

Tube # [OH-] pOH pH bromothymol blue red cabbage extract alizarin yellow phenolphtalein 1 dark blue yellow/brown dark red purple 2 dark blue yellow/brown red/brown purple 3 dark blue light blue/green yellow/brown purple 4 dark blue light blue yellow purple 5 dark blue blue yellow purple 6 dark blue dark blue yellow clear 7 light blue dark blue yellow clear

Explanation / Answer

For 0.1M NaOH solution

[OH-] = NaOH molarity= 0.1 M

pOH = -log[OH-] =-log(0.1) = 1

pH = 14- pOH = 14- 1 = 13

Tube 1: (0.1M Sodium Hydroxide)(5mL) = (?)(10mL) = 0.05M 10ml solution

(?) = (0.1M Sodium Hydroxide)(5mL) / (10mL) = 0.05M

[OH-] = NaOH molarity= 0.05 M

pOH = -log[OH-] =-log(0.05) = 1.3

pH = 14- pOH = 14- 1 = 12.7

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