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Enter your answer in the provided box. what is the original molarity of an aqueous solution of ammonia (nh3) whose ph is 11. 02 at 25°c? (kb for nh3 = 1. 8 × 10−5) m

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 .

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