回樓上
[H+]=根號(0.25x10-8.69)
-log[H+]=根號(log22 +log108.69 )
-log[H+]=log2+log104.35
-log[H+]=0.3010+4.35=4.65
燒瓶內有 0.25 M HBrO(aq),問其 pH 值是多少?【(pKa = 8.69),(log2 = 0.3010)】Ka=[H+]*[A-] /[HA][A-]*[H+]=Ka*[HA] HA-------------> [H+]+[A-] 單質子酸解離平衡下[A-]=[H+]故得[H+]2=Ka*[HA][H+]=(Ka*[HA])1/2pH =-log[H+]=-log(Ka*[HA])1/2 =-log(Ka*0.25 )1/2 =pKa/2+log2 =4.345+ 0.3010=4.64...
燒瓶內有 0.25 M HBrO(aq),問其 pH 值是多少?【(pKa = 8.69),(log2 = 0.3010)】Ka=[H+]*[A-] /[HA][A-]*[H+]=Ka*[HA] HA-------------> [H+]+[A-] 單質子酸解離平衡下[A-]=[H+]故得[H+]2=Ka*[HA][H+]=(Ka*[HA])1/2pH =-log[H+]=-log(Ka*[HA])1/2 =-log(Ka*0.25 )1/2 =pKa/2+log2 =4.345+ 0.3010=4.646
50.燒瓶內有 0.25 M HBrO(aq),問其 pH 值是多少?【(pKa..-阿摩線上測驗