Question

15£®£¨1£©ÒÑÖª£ºa³£ÎÂÏ£¬´×ËáºÍNH₃•H₂OµÄµçÀë³£ÊýÏàͬ£®
b£®CH₃COOH+NaHCO₃¨TCH₃COONa+CO₂¡ü+H₂O£¬Ôò
¢ÙCH₃COONH₄ÈÜÒº³ÊÖÐÐÔ£¨Ìî¡°Ëá¡°¡¢¡°¼î¡°»ò¡°ÖС°£¬ÏÂͬ£©£®
¢ÚNH₄HCO₃ÈÜÒº³Ê¼îÐÔ£¬ÈÜÒºÖÐÎïÖʵÄÁ¿Å¨¶È×î´óµÄÀë×ÓÊÇNH₄⁺£¨Ìѧʽ£©£®
£¨2£©100¡æʱ£¬Kw=1.0¡Á10^-12£¬¸ÃζÈϲâµÃ0.1mol/L Na₂AÈÜÒºµÄpH=6£®
¢ÙH₂AÔÚË®ÈÜÒºÖеĵçÀë·½³ÌʽΪH₂A¨T2H⁺+A^2-£»
¢Ú¸ÃζÈÏ£¬½«0.01mol/L H₂A ÈÜҺϡÊÍ20±¶ºó£¬ÈÜÒºµÄpH=3£»
¢ÛÌå»ýÏàµÈ¡¢pH¾ùΪ1µÄÑÎËáÓëH₂AÈÜÒº·Ö±ðÓë×ãÁ¿µÄZn·´Ó¦£¬²úÉúµÄÇâÆøC£»
A£®ÑÎËá¶à   B£®H₂A¶à   C£®Ò»Ñù¶à   D£®ÎÞ·¨È·¶¨
¢Ü½«0.1mol/L H₂AÈÜÒºÓë0.2mol/L°±Ë®µÈÌå»ý»ìºÏ£¬ÍêÈ«·´Ó¦ºóÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨NH₄⁺£©£¾c£¨A^2-£©£¾c£¨H⁺£©£¾c£¨OH^-£©£»
£¨3£©ÔÚijζÈʱ£¬²âµÃ0.01mol•L^-1µÄNaOHÈÜÒºµÄpH=10
¢Ù¸ÃζÈÏÂË®µÄÀë×Ó»ý³£ÊýK_w=1¡Á10^-12£®
¢ÚÔÚ´ËζÈÏ£¬½«pH=aµÄNaOHÈÜÒºV_a LÓëpH=bµÄÁòËáV_b L»ìºÏ£¨ºöÂÔÈÜÒºÌå»ý±ä»¯£©£®
ÈôËùµÃ»ìºÏҺΪÖÐÐÔ£¬ÇÒa=12£¬b=2£¬ÔòV_a£ºV_b=100£º1
ÈôËùµÃ»ìºÏÒºpH=10£¬ÇÒa=12£¬b=2£¬ÔòV_a£ºV_b=2£º99
£¨4£©ÔÚ25¡æʱ£¬pHµÈÓÚ9ºÍpHµÈÓÚ11µÄÁ½ÖÖÇâÑõ»¯ÄÆÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºÖеÄc£¨H⁺£©=2¡Á10^-11mol/L £¨lg2=0.3£¬lg5=0.7£©

Answer 1

·ÖÎö £¨1£©¢ÙÏàͬζÈÏ£¬µçÀëƽºâ³£ÊýÏàͬµÄÈõµç½âÖÊ£¬ÆäµçÀë³Ì¶ÈÏàµÈ£»
¢Ú¸ù¾ÝÇ¿ËáÖÆÈ¡ÈõËáÅжϴ×ËáºÍ̼ËáµÄÏà¶ÔÇ¿Èõ£»
£¨2£©99¡æʱ£¬K_w=1.0¡Á10^-12£¬¸ÃζÈϲâµÃ0.1mol•L^-1 Na₂AÈÜÒºµÄpH=6£¬ËµÃ÷Na₂AÊÇÇ¿ËáÇ¿¼îÑΣ»
¢ÙH₂AÊÇÇ¿ËáÍêÈ«µçÀ룻
¢Ú¸ù¾ÝÏ¡ÊÍÇ°ºóÈÜÖʵÄÎïÖʵÄÁ¿ÏàµÈ¼ÆË㣻
¢ÛËáÓë×ãÁ¿µÄп·´Ó¦Éú³ÉÇâÆøµÄÁ¿ÓëÇâÀë×ÓµÄÎïÖʵÄÁ¿³ÉÕý±È£»
¢Ü½«0.1mol•L^-1 H₂AÈÜÒºÓë0.2mol•L^-1°±Ë®µÈÌå»ý»ìºÏ£¬¶þÕßÇ¡ºÃ·´Ó¦Éú³É£¨NH₄⁺£©₂A£¬£¨NH₄⁺£©₂AÊÇÇ¿ËáÈõ¼îÑΣ¬ÆäÈÜÒº³ÊËáÐÔ£¬½áºÏµçºÉÊغãÈ·¶¨Àë×ÓŨ¶È´óС£»
£¨3£©¢Ù0.01mol/LµÄNaOHÈÜÒºµÄpHΪ10£¬Ôòc£¨H⁺£©=1¡Á10^-10mol/L£¬c£¨OH^-£©=0.01mol/L£¬¸ù¾ÝK_w=c£¨H⁺£©¡Ác£¨OH^-£©¼ÆË㣻
¢Ú»ìºÏÈÜÒº³ÊÖÐÐÔ£¬ÇÒÁòËáºÍÇâÑõ»¯Äƶ¼ÊÇÇ¿µç½âÖÊ£¬ËùÒÔËáÖÐÇâÀë×ÓÎïÖʵÄÁ¿ºÍ¼îÖÐÇâÑõ¸ùÀë×ÓµÄÎïÖʵÄÁ¿ÏàµÈ£»
£¨4£©ÏȼÆËã»ìºÏÈÜÒºÖÐc£¨OH^-£©£¬ÔÙ¸ù¾ÝKw=c£¨OH^-£©£¬c£¨H⁺£©¼ÆËãc£¨H⁺£©£®

½â´ð ½â£º£¨1£©¢Ù¸ù¾ÝÌâ¸øÐÅÏ¢Öª£¬ÏàͬÌõ¼þÏ£¬´×ËáºÍһˮºÏ°±µÄµçÀëƽºâ³£ÊýÏàµÈ£¬Ôò¶þÕߵĵçÀë³Ì¶ÈÏàµÈ£¬´×Ëáï§ÖÐ笠ùÀë×Ӻʹ×Ëá¸ùÀë×ÓË®½â³Ì¶ÈÏàµÈ£¬ËùÒÔ´×Ëáï§ÈÜÒº³ÊÖÐÐÔ£»
¹Ê´ð°¸Îª£ºÖУ»
¢Ú̼ËáÇâ¸ùÀë×ÓË®½â³Ì¶È´óÓÚ笠ùÀë×Ó£¬ÈÜÒºÏÔ¼îÐÔ£¬ËùÒÔÈÜÒºÖÐÀë×ÓŨ¶È×î´óµÄÊÇNH₄⁺£¬
¹Ê´ð°¸Îª£º¼îÐÔ£»NH₄⁺£»
£¨2£©99¡æʱ£¬K_w=1.0¡Á10^-12£¬¸ÃζÈÏ´¿Ë®µÄpH=6£¬²âµÃ0.1mol•L^-1 Na₂AÈÜÒºµÄpH=6£¬ËµÃ÷Na₂AÊÇÇ¿ËáÇ¿¼îÑΣ¬
¢ÙH₂AÊÇÇ¿ËᣬÔÚË®ÈÜÒºÀïÍêÈ«µçÀ룬ÆäµçÀë·½³ÌʽΪH₂A¨T2H⁺+A^2-£¬
¹Ê´ð°¸Îª£ºH₂A¨T2H⁺+A^2-£»
¢Ú¸ÃζÈÏ£¬½«0.01mol•L^-1 H₂AÈÜҺϡÊ͵½20±¶ºó£¬ÈÜÒºÖÐÇâÀë×ÓŨ¶È=0.01mol/L¡Á2¡Â20=0.001mol/L£¬
ËùÒÔpH=3£¬
¹Ê´ð°¸Îª£º3£»¡¡
¢ÛËáÓë×ãÁ¿µÄп·´Ó¦Éú³ÉÇâÆøµÄÁ¿ÓëÇâÀë×ÓµÄÎïÖʵÄÁ¿³ÉÕý±È£¬Ìå»ýÏàµÈ¡¢pH=1µÄÑÎËáÓëH₂AÈÜÒºÖÐÇâÀë×ÓµÄÎïÖʵÄÁ¿ÏàµÈ£¬ËùÒÔÓë×ãÁ¿Ð¿ÍêÈ«·´Ó¦Ê±Éú³ÉµÄÇâÆøµÄÁ¿Ò»Ñù¶à£¬
¹Ê´ð°¸Îª£ºC£»
¢Ü½«0.1mol•L^-1 H₂AÈÜÒºÓë0.2mol•L^-1°±Ë®µÈÌå»ý»ìºÏ£¬¶þÕßÇ¡ºÃ·´Ó¦Éú³É£¨NH₄⁺£©₂A£¬£¨NH₄⁺£©₂AÊÇÇ¿ËáÈõ¼îÑΣ¬ÆäÈÜÒº³ÊËáÐÔ£¬ËùÒÔc£¨H⁺£©£¾c£¨OH^-£©£¬ï§¸ùÀë×ÓÄÜË®½âµ«½Ï΢Èõ£¬¸ù¾ÝµçºÉÊغãµÃc£¨H⁺£©+c£¨NH₄⁺£©=2c£¨A^2-£©+c£¨OH^-£©£¬ËùÒÔc£¨NH₄⁺£©£¾c£¨A^2-£©£¬Ë®µÄµçÀë½Ï΢Èõ£¬ËùÒÔc£¨A^2-£©£¾c£¨H⁺£©£¬ÔòÀë×ÓŨ¶È´óС˳ÐòÊÇc£¨NH₄⁺£©£¾c£¨A^2-£©£¾c£¨H⁺£©£¾c£¨OH^-£©£¬
¹Ê´ð°¸Îª£ºc£¨NH₄⁺£©£¾c£¨A^2-£©£¾c£¨H⁺£©£¾c£¨OH^-£©£®
£¨3£©¢Ù0.01mol/LµÄNaOHÈÜÒºµÄpHΪ10£¬Ôòc£¨H⁺£©=1¡Á10^-10mol/L£¬c£¨OH^-£©=0.01mol/L£¬K_w=c£¨H⁺£©¡Ác£¨OH^-£©=1¡Á10^-12£¬
¹Ê´ð°¸Îª£º1¡Á10^-12£»
¢Ú»ìºÏÈÜÒº³ÊÖÐÐÔ£¬ÇÒÁòËáºÍÇâÑõ»¯Äƶ¼ÊÇÇ¿µç½âÖÊ£¬ËùÒÔËáÖÐÇâÀë×ÓÎïÖʵÄÁ¿ºÍ¼îÖÐÇâÑõ¸ùÀë×ÓµÄÎïÖʵÄÁ¿ÏàµÈ£¬pH=12µÄNaOHÈÜÒºÖÐÇâÑõ¸ùÀë×ÓŨ¶È=1mol/L£¬pH=2µÄH₂SO₄ÈÜÒºÇâÀë×ÓŨ¶È=10^-2 mol/L£¬1mol/L¡ÁV_aL=10^-2 mol/L¡ÁV_bL£¬ÔòV_a£ºV_b=100£º1£»
ÈôËùµÃ»ìºÏÒºpH=10£¬pH=12µÄNaOHÈÜÒºÖÐÇâÑõ¸ùÀë×ÓŨ¶È=1mol/L£¬pH=2µÄH₂SO₄ÈÜÒºÇâÀë×ÓŨ¶È=10^-2 mol/L£¬$\frac{1¡Á{V}_{a}-1{0}^{-2}¡Á{V}_{b}}{{V}_{a}+{V}_{b}}$=10^-2 ÔòV_a£ºV_b=2£º99£¬
¹Ê´ð°¸Îª£º100£º1£»2£º99£»
£¨4£©pH=9µÄÈÜÒºÖÐc£¨OH^-£©=$\frac{1{0}^{-14}}{1{0}^{-9}}$mol/L=10^-5mol/L£¬pH=11µÄÈÜÒºÖÐc£¨OH^-£©=$\frac{1{0}^{-14}}{1{0}^{-11}}$mol/L=10^-3mol/L£¬ÔòµÈÌå»ý»ìºÏÈÜÒºÖÐc£¨OH^-£©=$\frac{1{0}^{-5}+1{0}^{-3}}{2}$mol/L£¬»ìºÏÈÜÒºÖÐc£¨H⁺£©=$\frac{1{0}^{-14}}{\frac{1{0}^{-5}+1{0}^{-3}}{2}}$mol/L¡Ö2¡Á10^-11mol/L£»
¹Ê´ð°¸Îª£º2¡Á10^-11mol/L£®

µãÆÀ ±¾Ì⿼²éÁËÈõµç½âÖʵĵçÀëƽºâ¡¢ÑÎÀàË®½â¡¢Ëá¼î»ìºÏµÄ¼ÆËãµÈ֪ʶµã£¬Ã÷È·Èõµç½âÖʵçÀëƽºâÌص㼰ÑÎÀàË®½âÌصãÊǽⱾÌâ¹Ø¼ü£¬¸ù¾ÝË®µÄÀë×Ó»ý³£Êý½øÐмÆËã¼´¿É£¬ÌâÄ¿ÄѶÈÖеȣ®