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What is the pH of a 300mL solution of 2M NAOH?

Respuesta :

muhammadssempijja707

explanation:

[NaOH] = 2M

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

[OH-]=[NaOH]

[OH-] = 2M

From Kw = [OH-][H+]

But Kw = 10^-14 mol^2dm^-9

[H+] = Kw / [OH-]

[H+] = 10^-14 ÷ 2M

[H+] = 5 × 10^-15

pH = -log[H+]

pH = -log 5 × 10^-15

pH = 14.3

SvetkaChem

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