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irina1246 [14]
3 years ago
5

Determine the ph of a 0.188 m nh3 solution at 25°c. the kb of nh3 is 1.76 × 10-5.

Chemistry
1 answer:
Ber [7]3 years ago
4 0
When NH3 is dissolved in water, it dissociates  partially producing NH4+ ions and OH- ions. It has an equation:
NH3 + H2O → NH4+ + OH- 

<span>We use the Kb expression to determine the [OH-] concentration,
</span>
<span>Kb = [NH4+] [OH-] /* [NH3] </span>

We can write NH4+ as OH- since they are of equal ratio. 
<span>(1.76*10^-5) = [OH-]² / 0.188 
</span><span>[OH-]² = 3.3088*10^-6 </span>
<span>[OH-] = 1.819*10^-3 </span>

We calculate for H+ concentration as follows:

<span>[H+] [OH-] = 10^-14 </span>
<span>[H+] = 10^-14 / [OH-] </span>
<span>[H+] = 10^-14 / (1.819*10^-3) </span>
<span>[H+] = 5.50*10^-12 </span>

<span>pH = -log [H+] </span>
<span>pH = -log (5.5*10^-12) </span>
<span>pH = 11.26</span>
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