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[pH" of an aqueous solution of "0.6MNH(3...

[pH" of an aqueous solution of "0.6MNH_(3)" and "0.4MNH_(4)CI" is "],[9.4(pK_(b)=4.74)." The new "pH" when "0.1MCa(OH)_(2)" solution is "],[" added to it."],[[" (A) "9.86,," ( "]]

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pH of an aqueous solution of 0.6M NH_(3) and 0.4M NH_(4)CI is 9.4 (pK_(b) = 4.74) . The new pH when 0.1M Ca(OH)_(2) solution is added to it.

pH of an aqueous solution of 0.6M NH_(3) and 0.4M NH_(4)Cl is 9.4 (pK_(b) = 4.74) . The new pH when 0.1M Ca(OH)_(2) solution is added to it.

Calculate the pH of 0.1 MNH_(3) solution.

pK_(b) of aq. NH_(3) is 4.74 hence pH of 0.01 M NH_(3) solution is

The pH of a buffer solution of 0.1 M NH_(4)OH [ pK_(a)=4.0] and 0.1 M NH_(4)Cl is

An aqueous solution at room temperature contains 0.1 M NH_(4)Cl and 0.01M NH_(4)OH (pK_(b)=5), the pH of the solution is :

An aqueous solution at room temperature contains 0.1 M NH_(4)Cl and 0.01M NH_(4)OH (pK_(b)=5), the pH of the solution is :

The pH of 0.2 M aqueous solution of NH_4Cl will be (pK_b of NH_4OH = 4.74, log 2 = 0.3)