ÌâÄ¿ÄÚÈÝ
ÄÜÕýÈ·±íʾÏÂÁз´Ó¦µÄÀë×Ó·½³ÌʽÊÇ
A£®NaNO2ÈÜÒºÖмÓÈëËáÐÔKMnO4ÈÜÒº£º2MnO4£+5NO2£+ 6H+ £½ 2Mn2+ + 5NO3£ + 3H2O
B£®Ì¼ËáÇâï§ÈÜÒºÖмÓÈë×ãÁ¿ÇâÑõ»¯±µÈÜÒº: NH4++HCO3£+2OH£ £½ CO32£+ NH3?H2O +H2O
C£®Fe(NO3)3ÈÜÒºÖмÓÈë¹ýÁ¿µÄHIÈÜÒº£º2Fe3+ + 2I£ £½ 2Fe 2+ + I2
D£®ÓöèÐԵ缫µç½âÈÛÈÚÂÈ»¯ÄÆ£º2Cl£+2H2O £½ Cl2¡ü+H2¡ü+2OH£
A
¡¾½âÎö¡¿
ÊÔÌâ·ÖÎö£ºA¡¢MnO4?ÔÚËáÐÔÌõ¼þϰÑNO2-Ñõ»¯ÎªNO3?£¬Àë×Ó·½³ÌʽΪ£º2MnO4£+5NO2£+ 6H+ £½ 2Mn2+ + 5NO3£ + 3H2O£¬¹ÊAÕýÈ·£»B¡¢Ì¼ËáÇâï§ÈÜÒºÖмÓÈë×ãÁ¿ÇâÑõ»¯±µÈÜÒº£¬Éú³ÉBaCO3³Áµí£¬¹ÊB´íÎó£»C¡¢Fe(NO3)3ÈÜÒºÖмÓÈë¹ýÁ¿µÄHIÈÜÒº£¬·¢ÉúµÄ·´Ó¦ÎªNO3?¡¢H+Ñõ»¯I?£¬¹ÊC´íÎó£»D¡¢ÓöèÐԵ缫µç½âÈÛÈÚÂÈ»¯ÄÆ£¬Àë×Ó·½³ÌʽΪ£º2Cl?+2Na+
Cl2¡ü+2Na£¬¹ÊD´íÎó¡£
¿¼µã£º±¾Ì⿼²éÀë×Ó·½³Ìʽ¡£
25 ¡æÊ±£¬µçÀëÆ½ºâ³£Êý£º
»¯Ñ§Ê½ | CH3COOH | H2CO3 | HClO |
µçÀëÆ½ºâ³£Êý | 1.8¡Á10£5 | K1 4.3¡Á10£7 K2 5.6¡Á10£11 | 3.0¡Á10£8 |
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÎïÖʵÄÁ¿Å¨¶ÈΪ0.1 mol/LµÄÏÂÁÐËÄÖÖÎïÖÊ£ºa.Na2CO3¡¢b.NaClO¡¢c.CH3COONa¡¢d.NaHCO3£»pHÓÉ´óµ½Ð¡µÄ˳ÐòÊÇ________________£»(Ìî±àºÅ)
£¨2£©³£ÎÂÏÂ0.1 mol/LµÄCH3COOHÈÜÒº¼ÓˮϡÊ͹ý³Ì£¬ÏÂÁбí´ïʽµÄÊý¾ÝÒ»¶¨±äСµÄÊÇ_____£»
A£®c(H£«) B£®c(H£«)/c(CH3COOH) C£®c(H£«)¡¤c(OH£) D£®c(OH£)/c(H£«)
£¨3£©Ìå»ýΪ10 mL pH£½2µÄ´×ËáÈÜÒºÓëÒ»ÔªËáHX·Ö±ð¼ÓˮϡÊÍÖÁ1 000 mL£¬Ï¡Ê͹ý³ÌpH±ä»¯Èçͼ£¬ÔòHXµÄµçÀëÆ½ºâ³£Êý________(Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)´×ËáµÄƽºâ³£Êý£¬ÀíÓÉÊÇ__________________£¬
![]()
Ï¡Êͺó£¬HXÈÜÒºÖÐË®µçÀë³öÀ´µÄc(H£«)________´×ËáÈÜÒºÖÐË®µçÀë³öÀ´µÄc(H£«)(Ìî¡°´óÓÚ¡±¡¢¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±)£¬ÀíÓÉÊÇ£º__________________£»
£¨4£©25 ¡æÊ±£¬CH3COOHÓëCH3COONaµÄ»ìºÏÈÜÒº£¬Èô²âµÃ»ìºÏÒºpH£½6£¬ÔòÈÜÒºÖÐc(CH3COO£)£c(Na£«)£½________¡£(Ìî׼ȷÊýÖµ)