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*~BEN~* asked in 科學化學 · 1 decade ago

weak acid HA?

50.0 ml of a 0.1M solution of weak acid HA was titra with a 0.1M solution of NAOH the KA for HA is 1.0X10負四次方

quetion:the ph of 0.1M solution of weak acid HA?

2 Answers

Rating
  • Jimmy
    Lv 5
    1 decade ago
    Favorite Answer

    如果是要求弱酸解離後的pH

    那題目中的NaOH滴定似乎就沒用處了

    或是你少列了NaOH的體積呢?

    .

    如果忽略題目中鹼滴定而只考慮弱酸HA的解離......

    .

    [HA]←→[H+] + [A-]令HA解離xM

    [HA]=0.1-x近似於0.1

    [H+]=x

    [A-]=x

    Ka=[H+]*[A-]/[HA]=x*x/0.1=1.0*10^(-4)

    x^2=10^(-5)

    x=3.16*10^(-3)=[H+]

    pH=-Log[H+]=3-Log(3.16)=3-0.4997=2.5003近似於2.5

    我只是改寫一下bdref43所述的,不選我也沒關係

    Source(s): 學藝不精的高中生
  • ?
    Lv 7
    1 decade ago

    HA --> H A- Ka=1.0*10-4=[H ][A-]/[HA]

    [H ]=[A-], [HA] [A-]=0.1, [HA]=0.1-[H ], 代入上式: [H ]2/(0.1-[H ])=1.0*10-4, 解得 [H ]=3.113*10-3 M, pH=-log[H+]=2.51

    2008-05-01 22:03:28 補充:

    加號都被刪了,空白處和H右上方有+.

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