Question

現(xiàn)稱取變質(zhì)的氫氧化鈉樣品10g，配制成溶液，再向其中加入CaCl₂溶液（反應(yīng)的化學(xué)方程式為：Na₂CO₃+CaCl₂ = CaCO₃↓+2NaCl）。反應(yīng)時(shí)CaCl₂溶液與沉淀的質(zhì)量關(guān)系如圖所示。請(qǐng)解決以下問(wèn)題：
(1)計(jì)算樣品中NaOH的質(zhì)量；
(2)計(jì)算所加CaCl₂溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)；
(3)恰好完全反應(yīng)時(shí)燒杯中溶液的pH______7（選填“＞”、“＜”或“=”）

Answer 1

(1)解：設(shè)樣品中碳酸鈉的質(zhì)量為X
             Na₂CO₃+CaCO₃=CaCO₃ +2NaCl
              106                        100

              X                            5g
              106/100=X/5g
             X=5.3g  該樣品中氫氧化鈉的質(zhì)量為10g-5.3g=4.7g
           答氫氧化鈉的質(zhì)量為4.7g
(2)解：設(shè)加入的氯化鈣溶液中的溶質(zhì)質(zhì)量y
            Na₂CO₃+CaCl₂=CaCO₃ +2NaCl
                              111     100
                              y            5g
                             111/100=y/5g
                        y=5.55g所加氯化鈣溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)為 5.55g/50g×100%=11.1%

           答：氯化鈣溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)為11.1%