Discover a wealth of knowledge at Westonci.ca, where experts provide answers to your most pressing questions. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
The original molarity of an aqueous solution of ammonia whose pH is 11. 02 at 25°C is 0.556 M
Calculation ,
Given : pH = 11.5
[tex][H^{+} ][/tex] = [tex]10^{-11.5}[/tex]
[tex][OH^{-} ][/tex] = [tex]K_{w} /[H^{+} ][/tex] = [tex]10^{-14}[/tex]/ [tex]10^{-11.5}[/tex] = [tex]10^{-2.5}[/tex] M
Kb = [tex][NH_{4}^{+} ][OH^{-} ]/[NH_{4} OH][/tex] = [tex](10^{-2.5})^{2} /C[/tex] = 1.8×[tex]10^{-5}[/tex] M
C = 1/1.8 = 0.566 M
So, molarity of an aqueous solution of ammonia is 0.566 M
pH ia a measure of hydrogen ion concentration , a measure of acidity or alkalinity of the solution .pH scale usually range from 0 to 14 Aqueous solutions at 25°C witha pH less rhan 7 are acidic , whereas , those with pH greater than 7 are basic or alkaline. pOH ia a measure of hydroxide ion concentration , a measure of alkalinity of the solution .
To learn more about molarity
https://brainly.com/question/8732513
#SPJ4
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
