Question

6£®ÓÃŨ¶ÈΪ0.5mol•L^-1µÄNaOHÈÜÒºµÎ¶¨50mL 0.01mol•L^-1Al₂£¨SO₄£©₃ÈÜÒº£¬²âµÃµ¼µçÄÜÁ¦Ëæ¼ÓÈëÈÜÒºµÄÌå»ý±ä»¯ÇúÏßÈçͼËùʾ£®ÏÂÁÐ˵·¨ÖÐÕýÈ·µÄÊÇ£¨¡¡¡¡£©

• A£® aµãÈÜÒºÖÐÀë×ÓŨ¶È´æÔÚ¹Øϵ£ºc£¨Al³⁺£©+c£¨H⁺£©¨Tc£¨SO₄^2-£©+c£¨OH^-£©
• B£® bµãÈÜÒºÖÐÓУºc£¨Na⁺£©¨Tc£¨SO₄^2-£©
• C£® cµã¶ÔÓ¦µÄÈÜÒºÏÔËáÐÔ
• D£® µÎ¶¨¹ý³ÌÖÐʼÖÕ´æÔÚ£ºc£¨SO₄^2-£©£¾c£¨AlO₂^-£©

Answer 1

·ÖÎö ÓÃŨ¶ÈΪ0.5mol/LµÄNaOHÈÜÒºµÎ¶¨50mL 0.01mol/LAl₂£¨SO₄£©₃ÈÜÒº£¬·¢ÉúµÄ·´Ó¦ÓУºAl³⁺+3OH^-¨TAl£¨OH£©₃¡ý£¬Al£¨OH£©₃+OH^-¨T[Al£¨OH£©₄]^-£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÔ½´ó£¬Àë×ÓµçºÉÔ½¸ß£¬ÈÜÒºµ¼µçÄÜÁ¦Ô½Ç¿£¬½áºÏÈÜÒºÖеÄÊغã˼Ïë·ÖÎö£®

½â´ð ½â£ºÓÃŨ¶ÈΪ0.5mol/LµÄNaOHÈÜÒºµÎ¶¨50mL 0.01mol/LAl₂£¨SO₄£©₃ÈÜÒº£¬·¢ÉúµÄ·´Ó¦ÓУºAl³⁺+3OH^-¨TAl£¨OH£©₃¡ý£¬Al£¨OH£©₃+OH^-¨T[Al£¨OH£©₄]^-£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÔ½´ó£¬Àë×ÓµçºÉÔ½¸ß£¬ÈÜÒºµ¼µçÄÜÁ¦Ô½Ç¿£¬Æðʼʱc£¨Al³⁺£©=0.02mol/L£¬
A£®aµãΪµÎ¶¨Æðµã£¬ÈÜÒºÖнö´æÔÚAl₂£¨SO₄£©₃£¬¸ù¾ÝµçºÉÊغ㣬3c£¨Al³⁺£©+c£¨H⁺£©=c£¨OH^-£©+2c£¨SO₄^2-£©£¬¹ÊA´íÎó£»
B£®bµãʱ£¬¼ÓÈëNaOHÈÜÒºµÄÌå»ýΪ6mL£¬ÈÜÒºÖÐc£¨Na⁺£©=$\frac{0.5¡Á6}{6+50}$=0.05mol/L£¬c£¨SO₄^2-£©=$\frac{0.01¡Á3¡Á50}{50+6}$=0.03mol/L£¬Ôòc£¨Na⁺£©¡Ùc£¨SO₄^2-£©£¬¹ÊB´íÎó£»
C£®cµãʱ¼ÓÈëNaOHÈÜÒºµÄÌå»ýΪ8mL£¬¸ù¾Ý·´Ó¦¹Øϵ£¬´ËʱÈÜÒºÖÐÉú³ÉNa[Al£¨OH£©₄]£¬ÈÜҺӦΪ¼îÐÔ£¬¹ÊC´íÎó£»
D£®´ÓÎïÁÏÊغã½Ç¶È·ÖÎö£¬c£¨Al³⁺£©£¼c£¨SO₄^2-£©£¬Al³⁺Íêȫת»¯ÎªAlO₂^-£¬ÒÀÈ»´æÔÚc£¨SO₄^2-£©£¾c£¨AlO₂^-£©£¬ËùÒԵζ¨¹ý³ÌÖÐʼÖÕ´æÔÚ£ºc£¨SO₄^2-£©£¾c£¨AlO₂^-£©£¬¹ÊDÕýÈ·£®
¹ÊÑ¡D£®

µãÆÀ ±¾Ì⿼²éµÎ¶¨Ô­Àí£¬Ã÷È··´Ó¦·½³Ìʽ£¬¸ù¾ÝµÎ¼ÓµÄÁ¿¼ÆË㣬ÀÎÀΰÑÎÕÈÜÒºµÄÊغã˼ÏëÊǽâÌâµÄ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ¬ÊÔÌâÓÐÖúÓÚÅàÑø·ÖÎöÄÜÁ¦ºÍ˼άÄÜÁ¦£®