网奉山E学乾 省级精品课程申报 甲基橙指示剂滴定到终点，又消耗HC为15.01mL,则此试样含有什么物质？其含量各为多 少？ 答案：NaOH与Na2CO3的混合物 Na,CO,(HCI)-V;(HCD)M(Na,CO,) 样品 0.1000×0.01501×106.0 =61.86% 0.2572 (NaOH)(HC)(HC)(HCI)M(NaOH) m(样品 01000×002105-001500x400-9,39g 0.2572 23.Calculate the volume of sodium hydroxide (0.125 0 mol-L)needed to titrate 25.00 mL sulphuric acid (0.085 00 mol-L). Answer: V(NaOH)=2xc(H,SO,)V(H,SO,) c(NaOH) -2×0.0850x25.0mL=34.0mL 0.1250 24.0.715 6g solid sample containing NazCO,NaHCO3 and impurities is titrated with 0.200 0 mol-L HCI solution.When phenolphthalein changes color,22.06 mL of HCI solution is consumed.When methyl orange changes color,24.08 mL of HCI solution is consumed again. Calculate the content of NazCOs and NaHCO3. Answer: (NaCO)=(HCI)-(HCI)-M(NaCO,) (sample) _0.200×0.0206x1060=6535% 0.7156 (NaHCo,(HC(C)(HCM(NaHcO sample) 0.2000×0.02408-0.0206)×84.0=4.74% 0.7156 25.1.00 0g solid sample containing is dissolved into 25.00mL0.3000 moLHCL.The solution is titrated with 0.1000mol-L NaOH standard solution.At the end point,25.00 mLNaOH standard solution is consumed.Calculate the content of NazBO-10H2O. Answer:Na,B,)H)(HC)(NaOH)-(NaOH)B) 2 m (sample)