怎么计算0.20摩尔每升的磷酸的ph
\u4e13\u4e1a\u603b\u78f7-\u4ef7\u683c\u516c\u9053-\u70ed\u9500\u8d27\u6e90-\u54c1\u79cd\u9f50\u5168 磷酸的电离常5261数K1=7.6x10^-3,K2=6.3x10^-8K3=4.4x10^-13K2
K3相对K1非常小
就按4102K1计算,设电离的1653H+浓度内为x
H3PO4
=可逆=H++H2PO4-x
x
x溶液中未电离的H3PO4分子浓度为0.2-x根据
电离常数
容计算式K1=cH+
x
cH2PO42-
/
cH3PO4=
x^2
/
0.2-x=7.6x10^-3解得H+浓度x=0.03537molL-PH=1.45
绛旓細纾烽吀鍒嗘瀽绾祿搴︽槸85%銆傜７閰哥殑瀵嗗害鏄1.69锛屾懇灏旇川閲忔槸98锛屽垎鏋愮函娴撳害鏄85%锛璁＄畻鍑1000姣崌纾烽吀鐨鎽╁皵鏁板氨鏄祿搴︼細C锛圚3PO4锛=1000*1.69*85%闄や互98=14.7mol姣L銆傜７閰告槸涓绉嶅父瑙佺殑鏃犳満閰革紝鍖栧寮忎负H3PO4锛屽垎瀛愰噺涓97.995锛岀７閰镐笉鏄撴尌鍙戯紝涓嶆槗鍒嗚В锛屽嚑涔庢病鏈夋哀鍖栨э紝鍏锋湁閰哥殑閫氭э紝鏄笁鍏...
绛旓細鍋囪鍩规涓鍗0.02M (20mM) 纾烽吀鐩愮紦鍐叉恫 pH = pKa2 + Log[鎽╁皵(K2HPO4)/鎽╁皵(KH2PO4)]7.0= 7.2 + Log[鎽╁皵(K2HPO4)/鎽╁皵(KH2PO4)]閭ｄ箞,(1)鎽╁皵(K2HPO4)/鎽╁皵(KH2PO4)=10^0.2 鍙堬紝锛2锛夋懇灏(K2HPO4)+ 鎽╁皵(KH2PO4)= 1 x 0.02 = 0.02 鎽╁皵 瑙ｈ仈绔嬫柟绋嬬粍锛1锛...
绛旓細鏍规嵁PH=1/2(PKa+Pc)PKa=2PH-Pc=8-P0.2=8-0.7=7.3
绛旓細1.鎹甈鍘熷瓙瀹堟亽璁＄畻闇纾烽吀鐨勭墿璐ㄧ殑閲 ---2鍗囨贩鍚堟恫涓惈P鍘熷瓙0.2mol ---闇瑕佺函纾烽吀0.2mol ---闇瑕4鎽╁皵/鍗嘓3PO4婧舵恫鐨勪綋绉:(0.2mol/4mol/L)*1000=50mL 2.鎹挔绂诲瓙瀹堟亽姹傞渶姘㈡哀鍖栭挔鐨勮川閲 ---2鍗囨贩鍚堟恫涓惈P鍘熷瓙0.2mol(鍓嶉潰鐨勮绠楃粨鏋)---2鍗囨贩鍚堟恫涓惈NaH2PO4鍜孨a2HPO4鐨勬荤墿璐ㄧ殑...
绛旓細濡傛灉閰嶅埗鐨勭７閰婧舵恫涓1鍗囷紝閭ｄ箞闇瑕佺函纾烽吀0.05脳98=4.9鍏嬶紝濡傛灉浣跨敤鐨勬槸85%椋熺敤绾х７閰革紝閭ｄ箞闇瑕佺７閰哥殑璐ㄩ噺鏄4.9/0.85=5.765鍏嬨傚嵆绉板彇5.765鍏85%鐨勭７閰革紝鍦ㄦ悈鎷岀殑鍚屾椂锛屽姞姘村畾瀹硅嚦涓鍗囧嵆鍙
绛旓細鍋囪閰嶅埗500姣崌 杩戜技涓1.32鍏 绉板彇1.32鍏纾烽吀姘簩閾靛浐浣擄紝婧朵簬500姣崌钂搁姘翠腑锛屾悈鎷屽潎鍖鍗冲彲
绛旓細纾烽吀鐨勫惈閲85%鏄釜杩戜技鍊硷紝璁＄畻杩囩▼涓簲璇ユ寜85%璁＄畻銆傜７閰哥殑瀵嗗害鏄1.69锛屾懇灏旇川閲忔槸98锛屼笉鐭ラ亾瀵嗗害鏃犳硶璁＄畻鍑烘懇灏旇川閲忥紝璁＄畻鍑1000姣崌纾烽吀鐨勬懇灏鏁板氨鏄祿搴︼細C(H3PO4)=1000*1.69*85%/98=14.7mol/L C(1/3H3PO4)=1000*1.69*85%/(1/3*98)=44mol/L ...
绛旓細鍋囪閰嶄竴鍗100mmol/l 锛堝嵆0.1mol锛弆锛塸h6.0鐨勭７閰缂撳啿娑诧細ph = pka2 + log([hpo42-]/[h2po4-])6.0 = 7.2 + log([hpo42-]/[h2po4-])= 7.2 + log(hpo42-鎽╁皵/h2po4-鎽╁皵)(hpo42-鎽╁皵/h2po4-鎽╁皵)=10^(-1.2)= 0.0631 hpo42-鎽╁皵 + h2po4-鎽╁皵 = 0.1 姹傝В寰...
绛旓細85%浠ヤ笂娴撶７閰稿湪20鎽勬皬搴︽椂鐨勬祿搴︺佸瘑搴﹀鐓ц〃濡備笂銆傚亣瀹氫綘閭ｇ摱娴撶７閰哥殑娴撳害鏄89%锛屾煡琛ㄥ緱89%娴撶７閰哥殑瀵嗗害鏄1.738鍏/姣崌锛屽嵆1738鍏/鍗囷紝宸茬煡纾烽吀鐨勬懇灏璐ㄩ噺鏄98鍏/鎽╋紝鍒89%娴撶７閰哥殑鎽╁皵娴撳害鏄 1738g/L / 98g/mol * 0.89 = 15.7839mol/L ...
绛旓細1銆佺‘瀹氭墍闇閰嶅埗鐨勬憾娑蹭綋绉紝鏍规嵁鎽╁皵娴撳害鍜屾憾娑蹭綋绉紝浣跨敤鍏紡璁＄畻鎵闇璐ㄩ噺銆2銆佸皢鎵闇璐ㄩ噺鐨勭７閰婧惰В鍦ㄨ冻澶熺殑婧跺墏涓紝鍙互浣跨敤钂搁姘存垨鍏朵粬閫傚悎鐨勬憾鍓傦紝纭繚鍏呭垎鎼呮媽婧惰В锛岀洿鍒扮７閰稿畬鍏ㄦ憾瑙ｃ3銆佸皢婧舵恫杞Щ鑷冲閲忕摱涓紝骞剁敤閫傚綋鐨勬憾鍓傝ˉ鍏呰嚦鐩爣浣撶Н锛100姣崌锛夛紝鍐嶆鍏呭垎鎼呮媽婧舵恫锛岀‘淇濈７閰稿潎鍖鍒嗗竷鍦ㄦ憾娑...
