ph=7.45的缓冲溶液80ml,所需0.1mol/l na2hpo4和kh2po4其中pka2 h3po4=7.2.求各需上述溶_数学解答

ph=7.45的缓冲溶液80ml,所需0.1mol/l na2hpo4和kh2po4
其中pka2 h3po4=7.2.求各需上述溶液多少

人气：336 ℃　时间：2020-06-22 15:45:43

解答

PH=PKa2-lg[H2PO4-]/[HPO42-]7.45=7.2-lg[H2PO4-]/[HPO42-]lg[H2PO4-]/[HPO42-]=-0.25[H2PO4-]/[HPO42-]=0.562需要0.1mol/l kh2po4溶液80×0.560/1.562=28.8毫升需要0.1mol/l na2hpo4溶液80×1/1.562=51.2毫升...