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kiruha [24]
4 years ago
7

Calculate the [OH-] and the pH of 0.035 M KOH.

Chemistry
1 answer:
DanielleElmas [232]4 years ago
5 0

Answer:

pH

= 12.54

OH- concentration 28.84

Explanation:

KOH dissociates into K+ and OH-. The ratio of K+ and OH- ion is 1:1

In any aqueous solution, the H3O+ and OH - must satisfy the following condition -

[ H_3O^+] [OH^-] = k_w

[ H_3O^+] = \frac{k_w}{ [OH^-]}

[H_3O^+]  = \frac{1 * 10^{-14}}{3.5 *10^{-2}[H_3O^+]  = 2.857 * 10^{-13} M

pH =

- log [ H_3O^+]\\- log [2.857 * 10^{-13}]

pH

= - [-12.54]\\= 12.54

pOH = 14 - pH

pOH

= 14 - 12.54\\= 1.46

OH concentration = antilog 1.46

OH- concentration 28.84

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Consider the balanced equation below. 4nh3 3o2 right arrow. 2n2 6h2o what is the mole ratio of nh3 to n2? 2:4 4:2 4:4 7:2
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