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工業化學分析題目 (急) 贈15點
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最佳解答:
Let the formula of MOBS be HX where H is hydrogen, then: Ka = [H+][X-]/[HX] pKa = pH - log {[X-]/[HX]} 7.48 = 7.4 - log {[X-]/[HX]} log {[X-]/[HX]} = -0.08 [X-]/[HX] = 0.832 So mole ratio of X- to HX should be 0.832 Initially, no. of moles of MOBS = 5/223.29 = 0.02239 Suppose that no. of moles of KOH needed = x, then No. of moles of HX = 0.02239 - x No. of moles of X- = x x/(0.02239 - x) = 0.832 x = 0.0102 So vol. of KOH needed = 0.0102/0.626 x 1000 = 16.24 mL
其他解答:F893957AEB7D5408
工業化學分析題目 (急) 贈15點
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Calculate how many milliliters of 0.626 M KOH should be added to 5.00g MOBS (pKa=7.48, formula mass 223.29) to give a pH of 7.40. (10)最佳解答:
Let the formula of MOBS be HX where H is hydrogen, then: Ka = [H+][X-]/[HX] pKa = pH - log {[X-]/[HX]} 7.48 = 7.4 - log {[X-]/[HX]} log {[X-]/[HX]} = -0.08 [X-]/[HX] = 0.832 So mole ratio of X- to HX should be 0.832 Initially, no. of moles of MOBS = 5/223.29 = 0.02239 Suppose that no. of moles of KOH needed = x, then No. of moles of HX = 0.02239 - x No. of moles of X- = x x/(0.02239 - x) = 0.832 x = 0.0102 So vol. of KOH needed = 0.0102/0.626 x 1000 = 16.24 mL
其他解答:F893957AEB7D5408
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