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Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that K_a=K_b=1.8xx10^(-3). |
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Answer» Solution :`h=sqrt(K_b)=sqrt((K_w)/(K_(a)K_(B)))=sqrt((1XX10^(-14))/(1.8xx10^(-6)xx1.8xx10^-6))` `=sqrt((1)/(1.8)xx10^(-4))=sqrt(0.5555xx10^(-4))` `0.7453xx10^(-2)` `pH=(1)/(2)pK_w+(1)/(2)pK_(a)-(1)/(2)pK_(b)` Given that `K_a=K_b=1.8xx10^(-5)` If `K_a=K_b`, then, `pK_(a)=pK_(b)` `therefore pH=1/2pK_(w)=1/2(14)=7`. |
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