ç¼å²æ¶²:pH=pKa±1
å æ¤pKa=6.8±1,å³å¯ç¨
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H2PO4- <==>HPO42- + H+
Ka2 =[H+]*[HPO42-]/[H2PO4-]
=约=[H+]*n(HPO42-)/n(H2PO4-)
n(HPO42-)/n(H2PO4-)=Ka2/[H+]=6.3*10^-8 /10^-6.80)=0.40
n(HPO42-) + n(H2PO4-)=0.1mol/L*0.100L=0.01mol
å æ¤
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¸äºæ°¢é¾ï¼Mr136.09ï¼ç
ç©è´¨çé=0.00714mol
è´¨é=0.972g
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¸æ°¢äºé¾ï¼Mr228.22ï¼çç©è´¨çé=0.00286mol
è´¨é=0.652g