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denis23 [38]
3 years ago
8

I'M IN A TEST RN PLS HELP!

Chemistry
1 answer:
gtnhenbr [62]3 years ago
8 0
I think its B have a good day
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An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the
serg [7]

Answer:

pH = 12.43

Explanation:

<em>...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it</em>.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

<em>Moles KOH:</em>

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

<em>Moles HN3:</em>

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43

6 0
3 years ago
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