作业帮化学的base and acid的几道题,1.A table wine has a pH of 3.40.What is the hydronium ion concentration of thewine?2.A 2.5 x 10-3 M solution of an acid HA has a pH of 3.80 at 25°C.What is thehydronium ion concentration of the solution?Calculate K
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化学的base and acid的几道题,1.A table wine has a pH of 3.40.What is the hydronium ion concentration of thewine?2.A 2.5 x 10-3 M solution of an acid HA has a pH of 3.80 at 25°C.What is thehydronium ion concentration of the solution?Calculate K
化学的base and acid的几道题,
1.A table wine has a pH of 3.40.What is the hydronium ion concentration of the
wine?
2.A 2.5 x 10-3 M solution of an acid HA has a pH of 3.80 at 25°C.What is the
hydronium ion concentration of the solution?Calculate Ka for the acid.
3.The ionisation constant of a very weak acid,HA,is 4.0 x 10-9.Calculate the
equilibrium concentrations of H3O+,A- and HA in a 0.040 M solution.
笔记丢了 查又查不到 那位大大帮帮忙 记得要解题的思路阿 俺基础不好
化学的base and acid的几道题,1.A table wine has a pH of 3.40.What is the hydronium ion concentration of thewine?2.A 2.5 x 10-3 M solution of an acid HA has a pH of 3.80 at 25°C.What is thehydronium ion concentration of the solution?Calculate K
1,pH=3.4,根据pH的定义
pH=-lg[H+]
那么lg[H+]=-3.4
用计算器可以得到[H+]=10^-3.4mol/L=3.98*10^-4mol/L
2,pH=3.8
同样的可以得到[H+]=10^-3.8=(1.58*10^-4)mol/L
写出电离式
HA==(可逆)==H(+)+A(-)
这样得到
[HA]=(2.5*10^-3)-(1.58*10^-4)mol/L
[A-]=(1.58*10^-4)mol/L
根据Ka的定义,可以算出
Ka=[A-][H+]/[HA]=1.066*10^-5
3,这个酸的Ka=4.0 x 10-9
现在这个酸的浓度是c=0.040 mol/L
这个时候√(c/Ka)>>500
可以采取近似计算
[H+]=√cKa=1.26*10^-5mol/L
而根据电离方程式HA==(可逆)==H(+)+A(-)
[H+]与[A-]相等
所以[A-]=1.26*10^-5mol/L
SORRY 虽然我对化学很感兴趣 但是好象我学历尚浅你这些东西我都还没接触过