ÌâÄ¿ÄÚÈÝ

7£®Ä³Î¶ÈÏ£¬ÒÑÖª´×ËáµÄµçÀë³£ÊýKa=1.8¡Á10-5£¬´×ËáÒøµÄÈܶȻý³£ÊýKsp=1.6¡Á10-3£¬Ì¼ËáÒøµÄÈܶȻý³£ÊýKsp=8.3¡Á10-12£¬ÔÚ²»¿¼ÂÇÑÎÀàË®½âµÄÇé¿öÏ£¬ÏÂÁÐÓйØËµ·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®¸ÃζÈÏ´×ËáÒøµÄÈܽâ¶ÈԼΪ0.668g
B£®¸ÃζÈϱ¥ºÍ´×ËáË®ÈÜÒºµÄpH=2.5-$\frac{1}{2}$lg1.8
C£®Ä³ÈÜÒºÖк¬ÓÐCH3COO-ºÍCO32-£¬Å¨¶È¾ùΪ0.01mol/L£¬ÏòÆäÖÐÖðµÎ¼ÓÈë0.01 mol/LµÄAgNO3ÈÜÒº£¬CH3COO-Àë×ÓÏȲúÉú³Áµí
D£®´×ËáÈÜÒº¼ÓˮϡÊͺó£¬ÈÜÒºÖе¼µçÁ£×ÓµÄÊýÄ¿¼õÉÙ

·ÖÎö A£®´×ËáÒøµÄÈܶȻý³£ÊýKsp=1.6¡Á10-3£¬Ôòc£¨CH3COOAg£©=c£¨Ag+£©=$\sqrt{Ksp}$£¬¿ÉÒÔÇó³ö1LÈÜÒºÖÐCH3COOAgµÄÖÊÁ¿£¬ÔÙÇóÈܽâ¶È£»
B£®ÈÜҺŨ¶Èδ֪£¬²»ÄÜÈ·¶¨±¥ºÍÈÜÒºµÄpH£»
C£®¸ù¾ÝKspÇó³ö±¥ºÍ´×ËáÒøÈÜÒººÍ±¥ºÍ̼ËáÒøÈÜÒºÖÐÒøÀë×ÓµÄŨ¶È£¬ÒøÀë×ÓŨ¶ÈСµÄÏÈÎö³ö³Áµí£»
D£®¼ÓˮϡÊÍ´Ù½ø´×ËáµÄµçÀ룮

½â´ð ½â£ºA£®´×ËáÒøµÄÈܶȻý³£ÊýKsp=1.6¡Á10-3£¬Ôòc£¨CH3COOAg£©=c£¨Ag+£©=$\sqrt{Ksp}$=0.04mol/L£¬ËùÒÔ1LÈÜÒºÖÐCH3COOAgµÄÖÊÁ¿Îª0.04mol¡Á167g/mol=6.68g£¬Éè´×ËáÒøµÄÈܽâ¶ÈΪx£¬Ôò$\frac{x}{100}=\frac{6.68}{1000-6.68}$£¬½âµÃx¡Ö0.668g£¬¹ÊAÕýÈ·£»
B£®ÌâÖÐÖ»ÖªµÀ¸ÃζÈÏ´×ËáµÄµçÀë³£Êý£¬ÈÜҺŨ¶Èδ֪£¬²»ÄÜÈ·¶¨±¥ºÍÈÜÒºµÄpH£¬¹ÊB´íÎó£»
C£®Ä³ÈÜÒºÖк¬ÓÐCH3COO-ºÍCO32-£¬Å¨¶È¾ùΪ0.01mol/L£¬ÏòÆäÖÐÖðµÎ¼ÓÈë0.01 mol/LµÄAgNO3ÈÜÒº£¬Ðγɴ×ËáÒø³Áµíʱ£¬c£¨Ag+£©=$\frac{1.6¡Á1{0}^{-3}}{0.01}$=0.16mol/L£¬ÐγÉ̼ËáÒø³ÁµíʱÈÜÒºÖÐc£¨Ag+£©=$\sqrt{\frac{8.3¡Á1{0}^{-12}}{0.01}}$=$\sqrt{8.3}$¡Á10-5mol/L£¬ËùÒÔCO32-ÏÈÎö³ö³Áµí£¬¹ÊC´íÎó£»
D£®¼ÓˮϡÊÍ´Ù½ø´×ËáµÄµçÀ룬ÈÜÒºÖÐÇâÀë×Ó¡¢´×Ëá¸ùÀë×ÓµÄÎïÖʵľùÔö´ó£¬ËùÒÔÈÜÒºÖе¼µçÁ£×ÓµÄÊýÄ¿Ôö¶à£¬¹ÊD´íÎó£®
¹ÊÑ¡A£®

µãÆÀ ±¾Ì⿼²éÁËÈõµç½âÖʵĵçÀëÆ½ºâ¡¢ÈÜÒºpH¼ÆËãÈܶȻý³£ÊýµÄÓйؼÆË㣬ÌâÄ¿ÄѶÈÖеȣ¬×¢Òâ°ÑÎÕÈõµçµç½âÖʵĵçÀëÒÔ¼°ÄÑÈݵç½âÖʵÄÈÜ½âÆ½ºâ£¬½áºÏÈܶȻýµÄ¼ÆË㹫ʽÅжÏÊÇ·ñÉú³É³Áµí£¬²àÖØÓÚ¿¼²éѧÉúµÄ·ÖÎöÄÜÁ¦ºÍ¼ÆËãÄÜÁ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø