Cho từ từ dung dịch Ba(OH)₂ 0,25M (V ml) vào X chứa 20,08 gam hỗn hợp gồm NaHCO₃ và BaCl₂ và theo dõi lượng kết tủa. Sự phụ thuộc của khối lượng kết tủa thu được vào giá trị V được biểu diễn ở đồ thị bên dưới

Giá trị của a + b gần nhất với

Đoạn 1:

HCO₃^– + OH^– → CO₃^2– + H₂O

Ba²⁺ + CO₃^2– → BaCO₃

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGceaqabeaacaWGUb % WaaSbaaSqaaiaadkeacaWGHbWaaeWaaeaacaWGpbGaamisaaGaayjk % aiaawMcaamaaBaaameaacaaIYaaabeaaaSqabaGccqGH9aqpcaaIWa % GaaiilaiaaikdacaaI1aGaamiEaiabgkziUkaad6gadaWgaaWcbaGa % amOqaiaadggacaWGdbGaam4tamaaBaaameaacaaIZaaabeaaaSqaba % GccqGH9aqpcaWGUbWaaSbaaSqaaiaad+eacaWGibWaaWbaaWqabeaa % cqGHsislaaaaleqaaOGaeyypa0JaaGimaiaacYcacaaI1aGaamiEaa % qaaiaadkeacaWGubGaamOtaiaadsfacaqGGaGaamOqaiaadggacqGH % sgIRcaWGUbWaaSbaaSqaaiaadkeacaWGHbGaam4qaiaadYgadaWgaa % adbaGaaGOmaaqabaaaleqaaOGaeyypa0JaaGimaiaacYcacaaIYaGa % aGynaiaadIhaaaaa!643F! \begin{gathered} {n_{Ba{{\left( {OH} \right)}_2}}} = 0,25x \to {n_{BaC{O_3}}} = {n_{O{H^ - }}} = 0,5x \hfill \\ BTNT{\text{ }}Ba \to {n_{BaC{l_2}}} = 0,25x \hfill \\ \end{gathered} \)

Đoạn 2:

Ba(OH)₂ + NaHCO₃ → BaCO₃ + NaOH + H₂O

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGobGaamyyaiaadIeacaWGdbGaam4tamaaBaaa % meaacaaIZaaabeaaaSqabaaaaa!3E15! \to {n_{NaHC{O_3}}}\) đoạn 2 = \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaamOBamaaBa % aaleaacaWGcbGaamyyamaabmaabaGaam4taiaadIeaaiaawIcacaGL % PaaadaWgaaadbaGaaGOmaaqabaaaleqaaaaa!3CDC! {n_{Ba{{\left( {OH} \right)}_2}}}\) đoạn 2 = = 0,25.1,5x = 0,375x

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGcbaGaeyOKH4Qaam % OBamaaBaaaleaacaWGobGaamyyaiaadIeacaWGdbGaam4tamaaBaaa % meaacaaIZaaabeaaaSqabaaaaa!3E15! \to {n_{NaHC{O_3}}}\) tổng = 0,375x + 0,5x = 0,875x

→ 0,25x.208 + 0,875x.84 = 20,08

→ x = 0,16 lít

Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGceaqabeaacaWGUb % WaaSbaaSqaaiaadkeacaWGHbWaaeWaaeaacaWGpbGaamisaaGaayjk % aiaawMcaamaaBaaameaacaaIYaaabeaaaSqabaGcqaaaaaaaaaWdbi % abg2da9iaabccacaaIWaGaaiilaiaaiwdacaWG4bGaaiOlaiaaicda % caGGSaGaaGOmaiaaiwdacaqGGaGaeyypa0JaaeiiaiaaicdacaGGSa % GaaGimaiaaikdaaeaapaGaeyOKH4QaamOBamaaBaaaleaacaWGcbGa % amyyaiaadoeacaWGpbWaaSbaaWqaaiaaiodaaeqaaaWcbeaakiabg2 % da9iaad6gadaWgaaWcbaGaam4taiaadIeadaahaaadbeqaaiabgkHi % TaaaaSqabaGccqGH9aqpcaaIWaGaaiilaiaaicdacaaI0aaaaaa!5ADF! \begin{gathered} {n_{Ba{{\left( {OH} \right)}_2}}} = {\text{ }}0,5x.0,25{\text{ }} = {\text{ }}0,02 \hfill \\ \to {n_{BaC{O_3}}} = {n_{O{H^ - }}} = 0,04 \hfill \\ \end{gathered} \)

→ a = 7,88

Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaaciGaaiaabeqaamaabaabaaGceaqabeaacaWGUb % WaaSbaaSqaaiaadkeacaWGHbWaaeWaaeaacaWGpbGaamisaaGaayjk % aiaawMcaamaaBaaameaacaaIYaaabeaaaSqabaGccqGH9aqpcaaIXa % GaaiilaiaaiwdacaWG4bGaaiOlaiaaicdacaGGSaGaaGOmaiaaiwda % cqGH9aqpcaaIWaGaaiilaiaaicdacaaI2aaabaGaamOqaiaadsfaca % WGobGaamivaiaabccacaqGcbGaaeyyaiabgkziUkaad6gadaWgaaWc % baGaamOqaiaadggacaWGdbGaam4tamaaBaaameaacaaIZaaabeaaaS % qabaGccqGH9aqpcaaIWaGaaiilaiaaicdacaaI2aGaey4kaSIaamOB % amaaBaaaleaacaWGcbGaamyyaiaadoeacaWGSbWaaSbaaWqaaiaaik % daaeqaaaWcbeaakiabg2da9iaaicdacaGGSaGaaGymaaaaaa!6301! \begin{gathered} {n_{Ba{{\left( {OH} \right)}_2}}} = 1,5x.0,25 = 0,06 \hfill \\ BTNT{\text{ Ba}} \to {n_{BaC{O_3}}} = 0,06 + {n_{BaC{l_2}}} = 0,1 \hfill \\ \end{gathered} \)

→ b = 19,7 → a + b = 27,58