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Alexus [3.1K]
3 years ago
10

ANSWER FAST AND WILL GIVE BRAINLIEST What is the pH of a 300mL solution of 2M NAOH?

Chemistry
2 answers:
Harlamova29_29 [7]3 years ago
8 0

Answer:

14.3

Explanation:

pOH = - log[OH-]

[OH-] = [NaOH] = 2M because NaOH strong base, so it dissociate completely.

pOH = -log[2]

pH + pOH = 14

pH = 14 - pOH = 14 -(-log2) = 14.3

Viefleur [7K]3 years ago
5 0

explanation:

[NaOH] = 2M

NaOH(aq) <======> Na+(aq) + OH-(aq)

[OH-]=[NaOH]

[OH-] = 2M

<em>From</em><em> </em><em>Kw</em><em> </em><em>=</em><em> </em><em>[OH-]</em><em>[</em><em>H</em><em>+</em><em>]</em>

<em>But</em><em> </em><em>Kw</em><em> </em><em>=</em><em> </em><em>1</em><em>0</em><em>^</em><em>-</em><em>1</em><em>4</em><em> </em><em>mol^2dm^-9</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>[H+] = Kw / [OH-]

[H+] = 10^-14 ÷ 2M

[H+] = 5 × 10^-15

pH = -log[H+]

pH = -log 5 × 10^-15

pH = 14.3

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