Question

14£®ÏÂÁÐÐðÊöÖв»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

• A£® 0.1mol•L^-1NH₄HSÈÜÒºÖÐÓУºc£¨NH₄⁺£©=c£¨HS^-£©+c£¨S^2-£©+c£¨H₂S£©
• B£® 25¡æʱ£¬½«a mol•L^-1µÄ°±Ë®Óë0.01mol•L^-1µÄÑÎËáµÈÌå»ý»ìºÏºó£¬c£¨NH₄⁺£©=c£¨Cl^-£©£¬ÔòNH₃•H₂OµÄµçÀë³£ÊýΪ$\frac{1{0}^{-9}}{a-0.01}$
• C£® µÈŨ¶ÈµÄHCNºÍNaCN»ìºÏÈÜÒºÖÐÓУº2c£¨Na⁺£©=c£¨CN^-£©+c£¨HCN£©
• D£® µÈpHµÄ¢Ù£¨NH₄£©₂SO₄ÈÜÒº ¢ÚNH₄HSO₄ÈÜÒº  ¢ÛNH₄ClÈÜÒºÖУ¬c£¨NH₄⁺£©µÄ´óС¹Øϵ£º¢Ù=¢Û£¾¢Ú

Answer 1

·ÖÎö A£®Èκεç½âÖÊÈÜÒºÖж¼´æÔÚÎïÁÏÊغ㣬¸ù¾ÝÎïÁÏÊغãÅжϣ»
B£®ÔÚ25¡æÏ£¬Æ½ºâʱÈÜÒºÖÐc£¨NH₄⁺£©=c£¨Cl^-£©=0.005mol/L£¬¸ù¾ÝÎïÁÏÊغãµÃc£¨NH₃£®H₂O£©=£¨0.5a-0.005£©mol/L£¬¸ù¾ÝµçºÉÊغãµÃc£¨H⁺£©=c£¨OH^-£©=10^-7mol/L£¬ÈÜÒº³ÊÖÐÐÔ£¬NH₃•H₂OµÄµçÀë³£ÊýK_b=$\frac{c£¨O{H}^{-}£©£®c£¨N{{H}_{4}}^{+}£©}{c£¨N{H}_{3}£®{H}_{2}O£©}$£»
C£®Èκεç½âÖÊÈÜÒºÖж¼´æÔÚÎïÁÏÊغ㣬¸ù¾ÝÎïÁÏÊغãÅжϣ»
D£®ÇâÀë×ÓÒÖÖÆ笠ùÀë×ÓË®½â£¬ÂÈ»¯ï§ºÍÁòËáï§ÖÐ笠ùÀë×ÓË®½â³Ì¶ÈÏàͬ£®

½â´ð ½â£ºA£®Èκεç½âÖÊÈÜÒºÖж¼´æÔÚÎïÁÏÊغ㣬¸ù¾ÝÎïÁÏÊغãµÃc£¨NH₄⁺£©+c£¨NH₃•H₂O£©=c£¨HS^-£©+c£¨S^2-£©+c£¨H₂S£©£¬ËùÒÔc£¨NH₄⁺£©£¼c£¨HS^-£©+c£¨S^2-£©+c£¨H₂S£©£¬¹ÊA´íÎó£»
B£®ÔÚ25¡æÏ£¬Æ½ºâʱÈÜÒºÖÐc£¨NH₄⁺£©=c£¨Cl^-£©=0.005mol/L£¬¸ù¾ÝÎïÁÏÊغãµÃc£¨NH₃£®H₂O£©=£¨0.5a-0.005£©mol/L£¬¸ù¾ÝµçºÉÊغãµÃc£¨H⁺£©=c£¨OH^-£©=10^-7mol/L£¬ÈÜÒº³ÊÖÐÐÔ£¬NH₃•H₂OµÄµçÀë³£ÊýK_b=$\frac{c£¨O{H}^{-}£©£®c£¨N{{H}_{4}}^{+}£©}{c£¨N{H}_{3}£®{H}_{2}O£©}$=$\frac{1{0}^{-7}¡Á5¡Á1{0}^{-3}}{0.5a-5¡Á1{0}^{-3}}$=$\frac{1{0}^{-9}}{a-0.01}$£¬¹ÊBÕýÈ·£»
C£®Èκεç½âÖÊÈÜÒºÖж¼´æÔÚÎïÁÏÊغ㣬¸ù¾ÝÎïÁÏÊغãµÃ2c£¨Na⁺£©=c£¨CN^-£©+c£¨HCN£©£¬¹ÊCÕýÈ·£»
D£®pHµÄ£¨NH₄£©₂SO₄ÓëNH₄ClÈÜÒºÖУ¬¶¼ÊÇÇ¿ËáÈõ¼îÑΣ¬¸ù¾ÝÈÜÒº³ÊµçÖÐÐÔ¿ÉÅж϶þÕßNH₄⁺Ũ¶ÈÏàµÈ£¬ÓÉÓÚNH₄HSO₄µçÀëʱ²úÉúH⁺ʹÈÜÒº³ÊËáÐÔ£¬NH₄⁺µÄË®½â±»ÒÖÖÆ£¬Òò´ËNH₄HSO₄ÖÐNH₄⁺µÄŨ¶ÈСÓÚ£¨NH₄£©₂SO₄£¬¹Ê´ð°¸Ó¦Îª£º¢Ù=¢Û£¾¢Ú£¬¹ÊDÕýÈ·£»
¹ÊÑ¡A£®

µãÆÀ ±¾Ì⿼²éÀë×ÓŨ¶È´óС±È½Ï£¬Îª¸ßƵ¿¼µã£¬Éæ¼°Èõµç½âÖʵĵçÀëºÍÑÎÀàË®½â£¬Ã÷È·ÈÜÒºÖÐÈÜÖʼ°ÆäÐÔÖÊÊǽⱾÌâ¹Ø¼ü£¬×¢ÒâÎïÁÏÊغãºÍµçºÉÊغãµÄÁé»îÔËÓã¬ÄѵãÊÇBÑ¡ÏîµÄ¼ÆË㣬ÌâÄ¿ÄѶÈÖеȣ®