ijζÈ(t ¡æ)ʱ£¬Ë®µÄÀë×Ó»ýΪKW£½1.0¡Á10£­13mol2¡¤L£­2£¬Ôò¸ÃζÈ(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)________25 ¡æ£¬ÆäÀíÓÉÊÇ_______________________________________¡£
Èô½«´ËζÈÏÂpH£½11µÄ¿ÁÐÔÄÆÈÜÒºa LÓëpH£½1µÄÏ¡ÁòËáb L»ìºÏ(Éè»ìºÏºóÈÜÒºÌå»ýµÄ΢С±ä»¯ºöÂÔ²»¼Æ)£¬ÊÔͨ¹ý¼ÆËãÌîдÒÔϲ»Í¬Çé¿öʱÁ½ÖÖÈÜÒºµÄÌå»ý±È£º
(1)ÈôËùµÃ»ìºÏҺΪÖÐÐÔ£¬Ôòa¡Ãb£½________£»´ËÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòÊÇ___________________________________¡£
(2)ÈôËùµÃ»ìºÏÒºµÄpH£½2£¬Ôòa¡Ãb£½________¡£´ËÈÜÒºÖи÷ÖÖÀë×ÓµÄŨ¶ÈÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòÊÇ__________________________________________¡£

¡¡´óÓÚ¡¡Ë®µÄµçÀëÊÇÎüÈȹý³Ì£¬KWËæζÈÉý¸ß¶øÔö´ó¡¡(1)10¡Ã1¡¡c(Na£«)£¾c(SO)£¾c(H£«)£½c(OH£­)»òc(Na£«)£½2c(SO42¡ª)£¾c(H£«)£½c(OH£­)¡¡(2)9¡Ã2
c(H£«)£¾c(SO42¡ª)£¾c(Na£«)£¾c(OH£­)

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÓÃʵÑéÈ·¶¨Ä³ËáHAÊÇÈõµç½âÖÊ¡£¼×¡¢ÒÒÁ½Í¬Ñ§µÄ·½°¸ÊÇ£º
¼×£º¢Ù³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄHAÅäÖÆ0.1 mol/LµÄHAÈÜÒº100 mL£»¢ÚÓÃpHÊÔÖ½²â³ö¸ÃÈÜÒºµÄpH£¬¼´ÄÜÖ¤Ã÷HAÊÇÈõµç½âÖÊ¡£
ÒÒ£º¢ÙÓÃÒÑÖªÎïÖʵÄÁ¿Å¨¶ÈµÄHAÈÜÒº¡¢ÑÎËᣬ·Ö±ðÅäÖÆpH£½1µÄÁ½ÖÖËáÈÜÒº¸÷100 mL£»¢Ú·Ö±ðÈ¡ÕâÁ½ÖÖÈÜÒº¸÷10 mL£¬¼ÓˮϡÊÍÖÁ100 ml£»¢Û¸÷È¡ÏàͬÌå»ýµÄÁ½ÖÖÏ¡ÊÍҺװÈëÁ½¸öÊÔ¹ÜÖУ¬Í¬Ê±¼ÓÈë´¿¶ÈÏàͬµÄпÁ££¬¹Û²ìÏÖÏ󣬼´¿ÉÖ¤Ã÷HAÊÇÈõµç½âÖÊ¡£
(1)¼×·½°¸ÖУ¬ËµÃ÷HAÊÇÈõµç½âÖʵÄÀíÓÉÊDzâµÃÈÜÒºµÄpH________1(Ìî¡°>¡±¡°<¡±»ò¡°£½¡±)£¬ÒÒ·½°¸ÖУ¬ËµÃ÷HAÊÇÈõµç½âÖʵÄÏÖÏóÊÇ________(ÌîÐòºÅ)¡£
a£®×°ÑÎËáµÄÊÔ¹ÜÖзųöH2µÄËÙÂÊ¿ì
b£®×°HAÈÜÒºµÄÊÔ¹ÜÖзųöH2µÄËÙÂÊ¿ì
c£®Á½¸öÊÔ¹ÜÖвúÉúÆøÌåµÄËÙÂÊÒ»Ñù¿ì
(2)ÇëÄãÆÀ¼ÛÒÒ·½°¸ÖÐÄÑÒÔʵÏÖÖ®´¦ºÍ²»Í×Ö®´¦________¡£
(3)ÇëÄãÔÙÌá³öÒ»¸öºÏÀí¶ø±È½ÏÈÝÒ×½øÐеķ½°¸(Ò©Æ·¿ÉÈÎÓÃ)£¬Çë±íÊöʵÑé·½°¸£º________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÌ¼ËáÄÆ(´¿¶ÈԼΪ98%)Öк¬ÓÐCa2£«¡¢Mg2£«¡¢Fe3£«¡¢Cl£­ºÍSO42¡ªµÈÔÓÖÊ£¬Ìá´¿¹¤ÒÕÁ÷³ÌÈçÏ£º

¢ñ.̼ËáÄƵı¥ºÍÈÜÒºÔÚ²»Í¬Î¶ÈÏÂÎö³öµÄÈÜÖÊÈçÏÂͼËùʾ£º

¢ò.ÓйØÎïÖʵÄÈܶȻýÈçÏ£º

ÎïÖÊ
CaCO3
MgCO3
Ca(OH)2
Mg(OH)2
Fe(OH)3
Ksp
4.96¡Á10£­9
6.82¡Á10£­6
4.68¡Á10£­6
5.61¡Á10£­12
2.64¡Á10£­39
 
»Ø´ðÏÂÁÐÎÊÌ⣺
(1)¼ÓÈëNaOHÈÜÒººó¹ýÂ˵õ½µÄÂËÔüÖÐÖ÷Òªº¬ÓÐ________(Ìîд»¯Ñ§Ê½)¡£25¡æʱ£¬Ïòº¬ÓÐMg2£«¡¢Fe3£«µÄÈÜÒºÖеμÓNaOHÈÜÒº£¬µ±Á½ÖÖ³Áµí¹²´æÇÒÈÜÒºµÄpH£½8 ʱ£¬c(Mg2£«)¡Ãc(Fe3£«)£½________¡£
(2)²Ù×÷XΪ________£¬ÆäζÈÓ¦¿ØÖÆÔÚ_____________________________________
(3)ÓÐÈË´Ó¡°ÂÌÉ«»¯Ñ§¡±½Ç¶ÈÉèÏ뽫¡°Ä¸Òº¡±ÑØÁ÷³ÌÖÐÐéÏßËùʾ½øÐÐÑ­»·Ê¹Óá£ÇëÄã·ÖÎöʵ¼Ê¹¤ÒµÉú²úÖÐÊÇ·ñ¿ÉÐÐ________£¬²¢ËµÃ÷ÀíÓÉ______________________________
________________________________________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ij»¯Ñ§Ñ§Ï°Ð¡×éΪÑо¿HA¡¢HBºÍMOHµÄËá¼îÐÔµÄÏà¶ÔÇ¿Èõ,Éè¼ÆÒÔÏÂʵÑé:³£ÎÂϽ«pH=2µÄÁ½ÖÖËáÈÜÒºHA¡¢HBºÍpH=12µÄMOH¼îÈÜÒº¸÷1 mL,·Ö±ð¼ÓˮϡÊ͵½1 000 mL,ÆäpHµÄ±ä»¯ÓëÈÜÒºÌå»ýµÄ¹ØϵÈçͼ,¸ù¾ÝËù¸øµÄÊý¾Ý,Çë»Ø´ðÏÂÁÐÎÊÌâ:

(1)HAΪ        Ëá,HBΪ        Ëá(Ìî¡°Ç¿¡±»ò¡°Èõ¡±)¡£
(2)Èôc=9,ÔòÏ¡ÊͺóµÄÈýÖÖÈÜÒºÖÐ,ÓÉË®µçÀëµÄÇâÀë×ÓŨ¶ÈµÄ´óС˳ÐòΪ        (ÓÃËá¡¢¼î»¯Ñ§Ê½±íʾ)¡£
(3)Èôc=9,½«Ï¡ÊͺóµÄHAÈÜÒººÍMOHÈÜҺȡµÈÌå»ý»ìºÏ,ÔòËùµÃÈÜÒºÖÐc(A-)Óëc(M+)µÄ´óС¹ØϵΪc(A-)     (Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)c(M+)¡£
(4)Èôb+c=14,ÔòMOHΪ     ¼î(Ìî¡°Ç¿¡±»ò¡°Èõ¡±)¡£½«Ï¡ÊͺóµÄHBÈÜÒººÍMOHÈÜҺȡµÈÌå»ý»ìºÏ,ËùµÃ»ìºÏÈÜÒºµÄpH     7(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

A¡¢B¡¢C¡¢D·Ö±ðΪNH4HSO4¡¢Ba(OH)2¡¢AlCl3¡¢Na2CO3 4ÖÖÎïÖÊÖеÄ1ÖÖ£¬ÈÜÓÚË®¾ùÍêÈ«µçÀ룬ÏÖ½øÐÐÈçÏÂʵÑ飺
¢Ù×ãÁ¿AÈÜÒºÓëBÈÜÒº»ìºÏ¹²ÈÈ¿ÉÉú³É³Áµí¼×ºÍ´Ì¼¤ÐÔÆøζÆøÌ壻
¢ÚÉÙÁ¿AÈÜÒºÓëCÈÜÒº»ìºÏ¿ÉÉú³É³ÁµíÒÒ£»
¢ÛAÈÜÒºÓëBÈÜÒº¾ù¿ÉÈܽâ³ÁµíÒÒ£¬µ«¶¼²»ÄÜÈܽâ³Áµí¼×¡£
Çë»Ø´ð£º
(1)AµÄ»¯Ñ§Ê½Îª________£»ÊÒÎÂʱ£¬½«pHÏàµÈµÄAÈÜÒºÓëDÈÜÒº·Ö±ðÏ¡ÊÍ10±¶£¬pH·Ö±ð±äΪaºÍb£¬Ôòa________b(Ìî¡°>¡±¡°£½¡±»ò¡°<¡±)¡£
(2)¼ÓÈÈÕô¸ÉCÈÜÒº²¢×ÆÉÕ£¬×îºóËùµÃ¹ÌÌåΪ________(Ìѧʽ)¡£
(3)CÈÜÒºÓëDÈÜÒº·´Ó¦µÄÀë×Ó·½³ÌʽΪ__________________________________________
(4)ÏòBÈÜÒºÖÐÖðµÎ¼ÓÈëµÈÌå»ý¡¢µÈÎïÖʵÄÁ¿Å¨¶ÈµÄNaOHÈÜÒº£¬µÎ¼Ó¹ý³ÌÖÐË®µÄµçÀëƽºâ½«________(Ìî¡°ÕýÏò¡±¡°²»¡±»ò¡°ÄæÏò¡±)Òƶ¯£»×îÖÕËùµÃÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ________________________________________________________________________¡£
(5)ÒÑÖª³Áµí¼×µÄKsp£½x¡£½«0.03 mol¡¤L£­1µÄAÈÜÒºÓë0.01 mol¡¤L£­1µÄBÈÜÒºµÈÌå»ý»ìºÏ£¬»ìºÏÈÜÒºÖÐËá¸ùÀë×ÓµÄŨ¶ÈΪ________(Óú¬xµÄ´úÊýʽ±íʾ£¬»ìºÏºóÈÜÒºÌå»ý±ä»¯ºöÂÔ²»¼Æ)¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Èõµç½âÖʵĵçÀëƽºâ¡¢ÑÎÀàµÄË®½âƽºâºÍÄÑÈÜÎïµÄÈܽâƽºâ¾ùÊôÓÚ»¯Ñ§Æ½ºâ¡£
ÒÑÖªH2AÔÚË®ÖдæÔÚÒÔÏÂƽºâ£ºH2A=H£«£«HA£­£¬HA£­H£«£«A2£­¡£
(1)³£ÎÂÏÂNaHAÈÜÒºµÄpH________(ÌîÐòºÅ£¬ÏÂͬ)£¬Ô­ÒòÊÇ_____________¡£
A£®´óÓÚ7¡¡¡¡¡¡¡¡¡¡¡¡      B£®Ð¡ÓÚ7      C£®µÈÓÚ7       D£®ÎÞ·¨È·¶¨
(2)ijζÈÏ£¬ÈôÏò0.1 mol¡¤L£­1µÄNaHAÈÜÒºÖÐÖðµÎµÎ¼Ó0.1 mol¡¤L£­1 KOHÈÜÒºÖÁÈÜÒº³ÊÖÐÐÔ(ºöÂÔ»ìºÏºóÈÜÒºµÄÌå»ý±ä»¯)¡£´Ëʱ¸Ã»ìºÏÈÜÒºÖеÄÏÂÁйØϵһ¶¨ÕýÈ·µÄÊÇ________¡£
A£®c(H£«)¡¤c(OH£­)£½1.0¡Á10£­14
B£®c(Na£«)£«c(K£«)£½c(HA£­)£«2c(A2£­)
C£®c(Na£«)>c(K£«)
D£®c(Na£«)£«c(K£«)£½0.05 mol¡¤L£­1
(3)ÒÑÖª³£ÎÂÏÂH2AµÄ¸ÆÑÎ(CaA)µÄ±¥ºÍÈÜÒºÖдæÔÚÒÔÏÂƽºâ£ºCaA(s)??Ca2£«(aq)£«A2£­(aq)¡¡¦¤H>0¡£ÈôҪʹ¸ÃÈÜÒºÖÐCa2£«Å¨¶È±äС£¬¿É²ÉÈ¡µÄ´ëÊ©ÓÐ________¡£
A£®Éý¸ßζȡ¡¡¡¡¡¡¡¡¡¡¡ B£®½µµÍζÈ
C£®¼ÓÈëNH4Cl¾§Ìå  D£®¼ÓÈëNa2A¹ÌÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ij¶þÔªËá(»¯Ñ§Ê½ÓÃH2B±íʾ)ÔÚË®ÖеĵçÀë·½³ÌʽÊÇH2B===H£«£«HB£­£»HB£­H£«£«B2£­¡£»Ø´ðÏÂÁÐÎÊÌâ¡£
(1)Na2BÈÜÒºÏÔ_____(Ìî¡°ËáÐÔ¡±¡¢¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±)£¬ÀíÓÉÊÇ_______(ÓÃÀë×Ó·½³Ìʽ±íʾ)¡£
(2)ÔÚ0.1 mol¡¤L£­1µÄNa2BÈÜÒºÖУ¬ÏÂÁÐÁ£×ÓŨ¶È¹ØϵʽÕýÈ·µÄÊÇ____¡£
A£®c(B2£­)£«c(HB£­)£«c(H2B)£½0.1 mol¡¤L£­1
B£®c(Na£«)£«c(OH£­)£½c(H£«)£«c(HB£­)
C£®c(Na£«)£«c(H£«)£½c(OH£­)£«c(HB£­)£«2c(B2£­)
D£®c(Na£«)£½2c(B2£­)£«2c(HB£­)
(3)ÒÑÖª0.1 mol¡¤L£­1 NaHBÈÜÒºµÄpH£½2£¬Ôò0.1 mol¡¤L£­1 H2BÈÜÒºÖеÄÇâÀë×ÓµÄÎïÖʵÄÁ¿Å¨¶È¿ÉÄÜ____0.11 mol¡¤L£­1(Ìî¡°<¡±¡¢¡°>¡±»ò¡°£½¡±)£¬ÀíÓÉÊÇ_____¡£
(4)0.1 mol¡¤L£­1 NaHBÈÜÒºÖи÷ÖÖÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ_______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¢ñ£®³£ÎÂÏ£¬½«Ä³Ò»ÔªËáHA(¼×¡¢ÒÒ¡¢±û¡¢¶¡´ú±í²»Í¬µÄÒ»ÔªËá)ºÍNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬Á½ÖÖÈÜÒºµÄÎïÖʵÄÁ¿Å¨¶ÈºÍ»ìºÏÈÜÒºµÄpHÈçϱíËùʾ£º

ʵÑé
񅧏
HAÎïÖʵÄÁ¿
Ũ¶È/(mol¡¤L£­1)
NaOHÎïÖʵÄÁ¿
Ũ¶È/(mol¡¤L£­1)
»ìºÏºóÈÜ
񼵀pH
¼×
0.1
0.1
pH£½a
ÒÒ
0.12
0.1
pH£½7
±û
0.2
0.1
pH>7
¶¡
0.1
0.1
pH£½10
£¨1£©´Ó¼××éÇé¿ö·ÖÎö£¬ÈçºÎÅжÏHAÊÇÇ¿ËỹÊÇÈõË᣿                    ¡£
£¨2£©ÒÒ×é»ìºÏÈÜÒºÖÐÁ£×ÓŨ¶Èc(A£­)ºÍc(Na£«)µÄ´óС¹Øϵ                   ¡£
A£®Ç°Õß´ó      B£®ºóÕß´ó        C£®Á½ÕßÏàµÈ           D£®ÎÞ·¨ÅжÏ
£¨3£©´Ó±û×éʵÑé½á¹û·ÖÎö£¬¸Ã»ìºÏÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ________________¡£
£¨4£©·ÖÎö¶¡×éʵÑéÊý¾Ý£¬Ð´³ö»ìºÏÈÜÒºÖÐÏÂÁÐËãʽµÄ¾«È·½á¹û(Ö»ÁÐʽ)£ºc(Na£«)£­c(A£­)£½________mol¡¤L£­1¡£
¢ò£®Ä³¶þÔªËá(»¯Ñ§Ê½ÓÃH2B±íʾ)ÔÚË®ÖеĵçÀë·½³ÌʽÊÇ£ºH2B=H£«£«HB£­¡¢HB£­H£«£«B2£­
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨5£©ÔÚ0.1 mol¡¤L£­1µÄNa2BÈÜÒºÖУ¬ÏÂÁÐÁ£×ÓŨ¶È¹ØϵʽÕýÈ·µÄÊÇ              
A£®c(B2£­)£«c(HB£­)£½0.1 mol¡¤L£­1
B£®c(B2£­)£«c(HB£­)£«c(H2B)£½0.1 mol¡¤L£­1
C£®c(OH£­)£½c(H£«)£«c(HB£­)
D£®c(Na£«)£«c(H£«)£½c(OH£­)£«c(HB£­)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑ֪ijÈÜÒºÖÐÖ»´æÔÚOH£­¡¢H+¡¢NH4+¡¢Cl£­ËÄÖÖÀë×Ó£¬Ä³Í¬Ñ§ÍƲâÆäÀë×ÓŨ¶È´óС˳ÐòÓÐÈçÏÂËÄÖÖ¹Øϵ£º

A£®c(NH4+)£¾c(Cl£­)£¾c(OH£­)£¾c(H+) B£®c(Cl£­)£¾c(H+)£¾c(NH4+)£¾c(OH£­)
C£®c(Cl£­)£¾c(NH4+)£¾c(H+)£¾c(OH£­) D£®c(NH4+)£¾c(OH£­)£¾c(Cl£­)£¾c(H+)
¢Ù ÈôΪNH4Cl ÈÜÒº£¬ÔòËÄÖÖÀë×ÓŨ¶ÈµÄÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ___________¡££¨ÌîÐòºÅ£©
¢Ú ÈôΪHCl ºÍNH4Cl »ìºÏÈÜÒº£¬ÔòËÄÖÖÀë×ÓŨ¶ÈµÄÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ_________£»
ÈôΪNH3¡¤ H2O ºÍNH4Cl »ìºÏÈÜÒº£¬ËÄÖÖÀë×ÓŨ¶ÈµÄÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ__________¡£
¢Û Èô¸ÃÈÜÒºÊÇÓÉÌå»ýÏàµÈµÄHClÈÜÒººÍ°±Ë®»ìºÏ¶ø³É£¬ÇÒÇ¡ºÃ³ÊÖÐÐÔ£¬Ôò»ìºÏÇ°c(HCl)  _____c(NH3¡¤H2O)£¨Ìî¡°>¡±¡¢¡°<¡±¡¢»ò¡°=¡±£¬ÏÂͬ£©£¬»ìºÏºóÈÜÒºÖÐc(OH£­)____c(NH4+)¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸