£¨ÊµÑé°à×ö£©ÔÚ25 mL 0.1 mol/L NaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol/L CH3COOHÈÜÒº£¬ÇúÏßÈçÏÂͼËùʾ£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½ÏÕýÈ·µÄ £¨ £©
A£®ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÒ»¶¨¶¼ÓÐ
c(Na+) > c(CH3COO£) > c(OH£) > c(H+)
B£®ÔÚBµã£¬a > 12.5£¬ÇÒÓÐ
c(Na+) = c(CH3COO£) = c(OH£) = c(H+)
C£®ÔÚCµã£ºc(CH3COO£) = c(Na+) > c(H+) > c(OH£)
D£®ÔÚDµã£ºc(CH3COO£) + c(CH3COOH) = 2c(Na+)
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨ÊµÑé°à×ö£©ÔÚ25 mL 0.1 mol/L NaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol/L CH3COOHÈÜÒº£¬ÇúÏßÈçÏÂͼËùʾ£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½ÏÕýÈ·µÄ £¨ £©
A£®ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÒ»¶¨¶¼ÓÐ
c(Na+) > c(CH3COO£) > c(OH£) > c(H+)
B£®ÔÚBµã£¬a > 12.5£¬ÇÒÓÐ
c(Na+) = c(CH3COO£) = c(OH£) = c(H+)
C£®ÔÚCµã£ºc(CH3COO£) = c(Na+) > c(H+) > c(OH£)
D£®ÔÚDµã£ºc(CH3COO£) + c(CH3COOH) = 2c(Na+)
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨ÊµÑé°à×ö£©ÔÚ25 mL 0.1 mol/L NaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol/L CH3COOHÈÜÒº£¬ÇúÏßÈçÏÂͼËùʾ£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½ÏÕýÈ·µÄ £¨ £©
A£®ÔÚA¡¢B¼äÈÎÒ»µã£¬ÈÜÒºÖÐÒ»¶¨¶¼ÓÐ
c(Na+) > c(CH3COO£) > c(OH£) > c(H+)
B£®ÔÚBµã£¬a> 12.5£¬ÇÒÓÐ
c(Na+) = c(CH3COO£) = c(OH£) = c(H+)
C£®ÔÚCµã£ºc(CH3COO£) = c(Na+) > c(H+) > c(OH£)
D£®ÔÚDµã£ºc(CH3COO£) + c(CH3COOH) = 2c(Na+)
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com