250ml 0.15m nh4cl与100ml0.20m nh3混合形成的溶液的ph值
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发布时间：2024-10-16 15:56
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热心网友
时间：2024-10-23 04:43
反应前：0.05L x 0.15mol/L = 0.0075mol MH3; 0.2mol/L x 0.05L = 0.01mol NH4Cl; 0.1mol/L x 0.001L = 0.0001mol HCl PKa (NH4Cl) = 10 PH = PKa + log [NH3] / [NH4+] = 10 + log (0.0075/0.01) = 10 - 0.12 = 9.875 NH3 + HCl --- NH4Cl 反应后：0.0074mol NH3 ; 0.0101mol NH4Cl PH = PKa + log [NH3] / [NH4+] = 10 + log (0.0074/0.0101) = 10 - 0.135 = 9.865
