henderson-hassebalch的中文翻译

Henderson-Hasselbalch
Henderson-Hasselbalchequation是化学中关于酸碱平衡的一个方程式。PH=PK+log中的log数值是多少？

英语翻译
209 Henderson Road,#03-08
Henderson Industrial Park
Singapore 159551
怎么翻译?为什么?请指教
还有一种写法是：
209 Henderson Road
#03-08 Henderson Industrial Park
Singapore 159551

more pratice
howie henderson is a college student in New York City.this weekend he plans to study for an exam,but he doesn't want to study.he just wants to have fun.On friday night he dances.On saturday morning he plays tennis with his friend keith.On saturday afternoon his friends call him,and they have a party in his friend Roger's home.on sunday he goes to see a movie.the name of the movie is"cloudy monday".howie doesn't like it.It is a comedy,and howie doesn't like comedy very much.
At last,on sunday night ,howie studies for the exam.he studies from six in the evening to two in the morning.on monday morning howie goes to school and takes the exam.he is tired and now he is sorry he didn't study.
1.Where does Howie live?
_____________________
2.What's Howie?
_______________
3.What does he do on friday night?
_________________________
4.how long does he study on sunday night?
__________________________________
5.how does he feel on monday morning/
__________________________

CCO是什么的缩写?
来自：The president and CCO Henderson,who became the CCO of General Motor since 2008,will assume the CEO.

以例题一和例题二展开对Henderson–Hasselbalch equation的疑问
例题一：
计算混合4ml 0.1mol/l醋酸与5ml 0.1mol/l的NaOAC时其pH值.
[CH3COOH] = (4/9) x 0.1 mol/l
[CH3COO- ] = (5/9) x 0.1 mol/l
pKa= 4.76
pH = 4.76 + log (5/4)= 4.76 + 0.097 = 4.86
例题二：
a计算含0.1M乙酸和0.1M乙酸钠的缓冲液的PH值.
b乙酸的pKa为4.7.加入0.05M NaOH后pH值为多少?
由HH eq 得：pH=4.7 + lg[0.1/0.1]= 4.7
加入0.05M NaOH后,未溶乙酸浓度降到0.05M ,而乙酸根浓度升高到0.15M,因此pH=4.7 + lg[0.15/0.05]= 5.18
下面是疑问部分（答疑请提供思路并指明以下分析哪里出了问题）：
平衡公式Ka=[A-]*[H+]/[HA] 中使用的是平衡后的未溶的HA的浓度和平衡后的共轭碱的浓度,Henderson–Hasselbalch equation:pH=pKa + log（[A-]/[HA]）,是根据平衡常数 Ka=[A-]*[H+]/[HA] 两边取对数得出的,因此,pH=pKa + log（[A-]/[HA]）相应的也应该使用平衡后的未溶的酸HA和平衡后的共轭碱A-的浓度代入公式求解.于是,则有
在物质的量浓度是C的弱酸溶液HA中,根据HA H+ + A- 因为酸HA的起始浓度是C,设平衡后[H+]=X ,那么平衡后未溶的HA的浓度应该是C-X ,生成的A-的浓度和H+浓度相等,即[A-]=X ,则Ka= X的平方/ C-X ；
然而,上面例题一中 [CH3COOH] = (4/9) x 0.1 mol/l相当于HA的起始浓度C,[CH3COO- ] = (5/9) x 0.1 mol/l相当于A-的起始浓度,它们都不是平衡后的浓度,为何例题一把一个起始酸的浓度和起始共轭碱的浓度代入pH=pKa + log（[A-]/[HA]）来求平衡后的pH?
例题二也是同样的问题；“a计算含0.1M乙酸和0.1M乙酸钠的缓冲液的PH值.”如果它的意思是“配备缓冲液时每一升溶液分别加了0.1mol乙酸和乙酸钠”那么算pH时不应该是pH=4.7 + lg[0.1/0.1]= 4.7 ,因为0.1M只是起始浓度,问题二b也因此无法计算；如果它的意思是“这个缓冲液平衡后乙酸和乙酸钠的浓度均是0.1M”,那么a问的算法pH=4.7 + lg[0.1/0.1]= 4.7则没有问题,但是如果承接这个意思往下到b问,加入0.05M NaOH后,未溶乙酸浓度会降到0.05M ,而乙酸根浓度升到0.15M ,这相当于 “在一升溶液加入0.05mol乙酸,同时加入0.15mol乙酸根,问平衡后的pH值”.因此,在例题二b问中加入0.05M NaOH后,得出的未溶解酸 0.05M和共轭碱0.15M都只相当于一个起始浓度,此时每升溶液中的0.05mol乙酸有一部分还要转化为乙酸根,同时,每升溶液中的0.15mol乙酸根有一部分还要转化为乙酸,当它们平衡后,再把它们平衡后的各自的浓度代入pH=pKa + log（[A-]/[HA]）才是在问题二b问中所求的“加入0.05M NaOH后”的pH值.而在例题二,则直接把0.05M和0.15M这两个浓度代入公式,为何?此时0.05M和0.15M只相当于一个酸和共轭碱的起始量,能代入公式求pH值的应该是它们平衡后的量.不是么?
总括来说,pH=pKa + log（[A-]/[HA]）明明公式里[A-]和[HA]指的是平衡后的浓度,为何例题一和例题二中均用一个酸和共轭碱的起始浓度代入计算?