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(1)0.05mol/LµÄNaAc ²é±í£ºKa(HAc)= 1.8¡Á10-5 ½â£º¡ßc/Kb = 0.05/(Kw/Ka) = 0.05¡Á1.8¡Á10-5/10-14 > 500 ÓÖ¡ßcKb = 0.05¡Á10-14/1.8¡Á10-5 = 2.8¡Á10-11 > 10Kw ¡à[OH-] =

= 5.27¡Á10-6

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(2)0.05mol/LµÄNH4Cl ²é±í£ºKb(NH3)= 1.8¡Á10-5 ½â£º¡ßc/Ka = 0.05/(Kw/Kb) = 0.05¡Á1.8¡Á10-5/10-14 > 500 ÓÖ¡ßcKa = 0.05¡Á10-14/1.8¡Á10-5 = 2.8¡Á10-11 > 10Kw ¡à[H+] = = 5.27¡Á10-6

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(3)0.05mol/LµÄH3BO3 ²é±í£ºKa(H3BO3)=5.7¡Á10-10 ½â£º¡ßc/Ka1 = 0.05/5.7¡Á10-10 > 500 ÓÖ¡ßcKa = 0.05¡Á5.7¡Á10-10 > 10Kw ¡à[H+] = = 5.34¡Á10-6

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